For the elliptic integral of first kind, $K(m)=int_0^{pi/2}frac{d heta}{sqrt{1-m^2sin^2 heta}} $, it is well-known that $K(m)$ can be expressed in what Chowla and Selberg call "finite terms" (i.e. algebraic numbers and a finite product of Gamma functions of rational values) whenever $ifrac{K(sqrt{1-m^2})}{K(m)}$ belongs to an imaginary quadratic field $mathbb Q(sqrt{d})$
(see Theorem 7 in S. Chowla and A. Selberg, [On Epstein's zeta function, J. reine angew. Math. 227, 86-110, 196][1]).
Examples for these so called **elliptic integral singular values** are given on [this Wolfram page][2] (with some small typos) and in the note of J.M. Borwein and I.J. Zucker, "Elliptic integral evaluation of the Gamma function at
rational values of small denominator," [IMA Journal on Numerical Analysis, 12 (1992), 519-
526][3].
See also what Tito Piezas has to say about this in his pleasant-to-read [Collection of Algebraic Identities][4].
The following question arises:
- **For these singular values, is there** (always, or, if not always: when?) **a polynomial $P(t)$ of degree 3 with *integer* coefficients such that $K(m)=cintlimits_{t_0}^inftydfrac{dt}{sqrt{P(t)}} $ with $cinmathbb Q$?**
(EDIT: After Noam Elkies' remark, introduced $t_0$, the biggest real zero of $P$, instead of $0$ as the lower limit. Only "complete" integrals make sense here.)
In particular for $d=-7$, we have by the [Carlson symmetric form][5] $$frac12intlimits_0^inftydfrac{dt}{sqrt{t(t+1)(t+frac{8+3sqrt 7}{16})}} =K(k_7)=dfrac1{7^{1/4}4pi}Gammaleft(dfrac17 ight)Gammaleft(dfrac27 ight)Gammaleft(dfrac47 ight),$$ on the other hand I have seen somewhere (I can't remember the reference) $$intlimits_0^inftydfrac{dt}{sqrt{t(t^2+21t+112)}} =dfrac1{4pisqrt{7}}Gammaleft(dfrac17 ight)Gammaleft(dfrac27 ight)Gammaleft(dfrac47 ight)=frac{K(k_7)}{7^{1/4}}.$$
I would like to get it straight at least for this example:
- **Can the polynomial in the first integral be transformed into one with integer coefficients? And is there any sort of relationship between both above polynomials?**
Note that the ratio of the discriminants of the two above polynomials is $-2^{24}cdot7^3$, and both of them do *not* yield affirmative answers, as the second one would have to be divided by $sqrt7$ to obtain $K(k_7)$ directly!
- **EDIT: Follow-up question: If $P(t)$ is an integer cubic polynomial such that $intlimits_{t_0}^inftydfrac{dt}{sqrt{P(t)}} $ (with $t_0$ its biggest real zero) can be written in "finite terms", is this value always an algebraic multiple of an elliptic integral singular value $K(m)$?**
[1]: http://www.ncbi.nlm.nih.gov/pmc/articles/PMC1063041/pdf/pnas01544-0031.pdf
[2]: http://mathworld.wolfram.com/EllipticIntegralSingularValue.html
[3]: http://carma.newcastle.edu.au/jon/Preprints/Books/EMA/Exercises/For%20others/K-beta.pdf
[4]: https://www.sites.google.com/site/tpiezas/0026
[5]: http://en.wikipedia.org/wiki/Carlson_symmetric_form
The flurry of comments did not yet produce an answer to the question
concerning the complete elliptic integrals
$$
I_1 := frac12
intlimits_0^inftydfrac{dt}{sqrt{t(t+1)(t+frac{8+3sqrt{7}}{16})}}
$$
and
$$
I_2 := intlimits_0^inftydfrac{dt}{sqrt{t(t^2+21t+112)}}.
$$
It turns out that (i) Yes, $I_1$ can be transformed to a complete
elliptic integral associated to a cubic with integer coefficients, and
(ii) The identity $I_1 = 7^{1/4} I_2$ can then be recovered via
a form of
<a href="http://en.wikipedia.org/wiki/Landen's_transformation">Landen's transformation</a>.
This could be surmised by calculating the $j$-invariants of the
corresponding elliptic curves
$$
y^2 = x(x+1)(x+frac{8+3sqrt{7}}{16})
phantom{cong} ext{and}phantom{cong}
y^2 = x(x^2+21x+112).
$$
The first $j$-invariant is $j_1 = 16581375 = 255^3$, which is rational,
so there's a linear change of variable that transforms
$x(x+1)(x+frac{8+3sqrt{7}}{16})$ to a polynomial with integer coefficients.
The second $j$-invariant is $j_2 = -3375 = -15^3
eq j_1$, so we can't
get immediately from $j_1$ to $j_2$. But $j_1$ and $j_2$ are still
related by a $2$-isogeny [indeed $j_1 = j(sqrt{-7})$ and
$j_2 = j((1+sqrt{-7})/2)$], so $I_1$ and $I_2$ are related by
a Landen transformation.
For (i), first translate $t$ by $(8+3sqrt{7})/16$ to get
$$
I_1 := frac12
intlimits_{frac{8+3sqrt{7}}{16}}^infty
dfrac{dt}{sqrt{t
igl(t-frac{8+3sqrt{7}}{16}igr)
igl(t-frac{-8+3sqrt{7}}{16}igr)
}} phantom{0}.
$$
Then observe that
$$
igl(t-frac{8+3sqrt{7}}{16}igr)
igl(t-frac{-8+3sqrt{7}}{16}igr)
= t^2 - frac{3sqrt{7}}{8} + frac{3^2 7 - 8^2}{16^2}
= t^2 - frac{6sqrt{7}}{16} - frac1{16^2}.
$$
Thus the change of variable $x = 16 sqrt{7} cdot t$ yields
$$
I_1 = 2 cdot 7^{1/4}
intlimits_{21+8sqrt{7}}^infty
dfrac{dx}{sqrt{x (x^2-42x-7)}}
$$
with the integrand having integer coefficients.
For (ii), we apply Landen's change of variable for elliptic integrals
$int dt/sqrt{t(t^2+at+b)}$ with $a,b>0$: if $x = (t^2+at+b)/t$ then
$dt/sqrt{t(t^2+at+b)} = dx/sqrt{x(x^2+Ax+B)}$ where $(A,B) = (-2a, a^2-4b)$.
Moreover the map $t mapsto (t^2+at+b)/t$ maps the interval $0 < t < infty$
to $a + 2sqrt{b} leq x < infty$, with each $x$ value arising twice
except for the left endpoint $x = a+2sqrt{b}$ which has a single preimage
$t = sqrt{b}$ of multiplicity $2$. Therefore
$$
intlimits_0^inftydfrac{dt}{sqrt{t(t^2+at+b)}} =
2 intlimits_{a+2sqrt{b}}^inftydfrac{dx}{sqrt{x(x^2-AX-B)}}.
$$
The $I_2$ coefficients $(a,b) = (21,112)$ yield $(A,B) = (-42,-7)$ and
$a + 2 sqrt{b} = 21 + 8 sqrt{7}$. Therefore $I_1 = 7^{1/4} I_2$ as desired.
来源：这里
For the first one,
$$egin{align}
int_0^{infty} (operatorname{sech}x)^{2s}dx \
&= int_0^{infty} (operatorname{sech}^2x)^{s-1}operatorname{sech}^2x\, dx \
&= int_0^{infty} (1- anh^2x)^{s-1}\,mathrm{d}( anh x)\
&= int_0^1 (1-x^2)^{s-1} mathrm{d}x\
&= frac12 int_0^1 (1-x)^{s-1} x^{-frac12} mathrm{d}x\
&= frac12 B(s,frac12)=frac{sqrt{pi}}{2}frac{Gamma(s)}{Gamma(frac12+s)}
end{align}$$
and so your conjecture is correct.
For the second integral, let $x=2t$:
$$egin{align}
int_0^{infty} frac{1}{(1+cosh x)^s}mathrm{d}x \
&= 2int_0^{infty} frac{1}{(1+cosh 2t)^s}mathrm{d}t \
&= 2int_0^{infty} frac{1}{(2cosh^2(t))^s}mathrm{d}t\
&= 2^{1-s} int_0^{infty} (operatorname{sech}t)^{2s}mathrm{d}t\
&= 2^{-s} B(s,frac12)=frac{sqrt{pi}}{2^s}frac{Gamma(s)}{Gamma(frac12+s)}
end{align}$$
Extending **@nospoon**'s idea, we notice that
$$ a + cosh 2x = (a+1)cosh^2 x (1 - b anh^2 x), qquad b =frac{a-1}{a+1}. $$
If $a > -1$, then $b < 1$ and using the substitution $u = anh^2 x$ we get
$$ int_{0}^{infty} frac{dx}{(a + cosh x)^{s}} \, dx = frac{1}{(a+1)^{s}} int_{0}^{1} frac{(1 - u)^{s-1}}{(1 - b u)^ssqrt{u}} \, du. $$
Making further substitution $v = frac{1-u}{1-bu}$, we have
$$ int_{0}^{infty} frac{dx}{(a + cosh x)^{s}} \, dx = frac{1}{(a+1)^{s}} int_{0}^{1} frac{v^{s-1}}{sqrt{(1-v)(1-bv)}} \, dv. $$
This is easily integrated when $b = 0$ or $b = -1$, each correspondingto $a = 1$ or $a = 0$, but I doubt that this integral has a nice closed form in general. For example, when $s = 1/2$, this becomes an elliptic integral and *Mathematica 11* gives
$$ int_{0}^{1} frac{dv}{sqrt{v(1-v)(1-bv)}} = 2 left[ frac{1}{sqrt{b}} K left( sqrt{ frac{1}{b}} ight) + i K left( sqrt{1-b} ight) ight], quad 0 < b < 1, $$
where $K(k)$ is the [*complete elliptic integral of the 1st kind*](https://en.wikipedia.org/wiki/Elliptic_integral#Complete_elliptic_integral_of_the_first_kind).
>$$I:=int_{0}^{infty}frac{ln{(x)}}{sqrt{x}\,sqrt{x+1}\,sqrt{2x+1}}mathrm{d}x.$$
After first multiplying and dividing the integrand by 2, substitute $x=frac{t}{2}$:
$$I=int_{0}^{infty}frac{2ln{(x)}}{sqrt{2x}\,sqrt{2x+2}\,sqrt{2x+1}}mathrm{d}x=int_{0}^{infty}frac{ln{left(frac{t}{2} ight)}}{sqrt{t}\,sqrt{t+2}\,sqrt{t+1}}mathrm{d}t.$$
Next, substituting $t=frac{1}{u}$ yields:
$$egin{align}
I
&=-int_{0}^{infty}frac{ln{(2u)}}{sqrt{u}sqrt{u+1}sqrt{2u+1}}mathrm{d}u\
&=-int_{0}^{infty}frac{ln{(2)}}{sqrt{u}sqrt{u+1}sqrt{2u+1}}mathrm{d}u-int_{0}^{infty}frac{ln{(u)}}{sqrt{u}sqrt{u+1}sqrt{2u+1}}mathrm{d}u\
&=-int_{0}^{infty}frac{ln{(2)}}{sqrt{u}sqrt{u+1}sqrt{2u+1}}mathrm{d}u-I\
implies I&=-frac{ln{(2)}}{2}int_{0}^{infty}frac{mathrm{d}x}{sqrt{x}sqrt{x+1}sqrt{2x+1}}.
end{align}$$
Making the sequence of substitutions $x=frac{u-1}{2}$, then $u=frac{1}{t}$, and finally $t=sqrt{w}$, puts this integral into the form of a beta function:
$$egin{align}
int_{0}^{infty}frac{mathrm{d}x}{sqrt{x}sqrt{x+1}sqrt{2x+1}}
&=int_{1}^{infty}frac{mathrm{d}u}{sqrt{u-1}sqrt{u+1}sqrt{u}}\
&=int_{1}^{infty}frac{mathrm{d}u}{sqrt{u^2-1}sqrt{u}}\
&=int_{1}^{0}frac{t^{3/2}}{sqrt{1-t^2}}frac{(-1)}{t^2}mathrm{d}t\
&=int_{0}^{1}frac{mathrm{d}t}{sqrt{t}\,sqrt{1-t^2}}\
&=frac12int_{0}^{1}frac{mathrm{d}w}{w^{3/4}\,sqrt{1-w}}\
&=frac12operatorname{B}{left(frac14,frac12
ight)}\
&=frac12frac{Gamma{left(frac12
ight)}Gamma{left(frac14
ight)}}{Gamma{left(frac34
ight)}}\
&=frac{pi^{3/2}}{2^{1/2}Gamma^2{left(frac34
ight)}}
end{align}$$
Hence,
$$I=-frac{ln{(2)}}{2}frac{pi^{3/2}}{2^{1/2}Gamma^2{left(frac34 ight)}}=-frac{pi^{3/2}\,ln{(2)}}{2^{3/2}\,Gamma^2{left(frac34 ight)}}.~~~lacksquare$$
---
**Possible Alternative:** You could also derive the answer from the complete elliptic integral of the first kind instead of from the beta function by making the substitution $t=z^2$ instead of $t=sqrt{w}$.
$$egin{align}
int_{0}^{infty}frac{mathrm{d}x}{sqrt{x}sqrt{x+1}sqrt{2x+1}}
&=int_{1}^{infty}frac{mathrm{d}u}{sqrt{u-1}sqrt{u+1}sqrt{u}}\
&=int_{1}^{infty}frac{mathrm{d}u}{sqrt{u^2-1}sqrt{u}}\
&=int_{1}^{0}frac{t^{3/2}}{sqrt{1-t^2}}frac{(-1)}{t^2}mathrm{d}t\
&=int_{0}^{1}frac{mathrm{d}t}{sqrt{t}\,sqrt{1-t^2}}\
&=2int_{0}^{1}frac{mathrm{d}z}{sqrt{1-z^4}}\
&=2\,K{(-1)}\
&=frac{Gamma^2{left(frac14
ight)}}{2sqrt{2pi}}\
&=frac{pi^{3/2}}{2^{1/2}Gamma^2{left(frac34
ight)}}.
end{align}$$
$$int_0^1frac{mathrm dx}{sqrt{1-x} sqrt[4]x sqrt[4]{2-x\,sqrt3}}stackrel?=frac{2\,sqrt2}{3\,sqrt[8]3}pi ag1$$
The equality numerically holds up to at least $10^4$ decimal digits.
> Can we prove that the equality is exact?
An equivalent form of this conjecture is
$$Ileft(frac{sqrt3}2; frac14,frac14
ight)stackrel?=frac23, ag2$$
where $Ileft(z; a,b
ight)$ is the [regularized beta function](http://mathworld.wolfram.com/RegularizedBetaFunction.html).
---
Even simpler case:
$$int_0^1frac{mathrm dx}{sqrt{1-x} sqrt[6]{9-x} sqrt[3]x}stackrel?=fracpi{sqrt3}, ag3$$
which is equivalent to
$$Ileft(frac19; frac16,frac13
ight)stackrel?=frac12. ag4$$
---
A [related question](https://math.stackexchange.com/questions/538564/conjecture-int-01-fracdx-sqrt3x-sqrt61-x-sqrt1-x-left-sqrt6).
For $alpha, eta, gamma in (0,1)$ satisfying $alpha+eta+gamma = 1$ and
$mu in mathbb{C} setminus [1,infty)$, define
$$
F_{alphaeta}(mu) = int_0^1frac{dx}{x^alpha(1-x)^eta(1-mu x)^gamma}
quad ext{ and }quad
Delta = frac{Gamma(1-alpha)Gamma(1-eta)}{Gamma(1+gamma)}
$$
When $|mu| < 1$, we can rewrite the integral $F_{alphaeta}(mu)$ as
$$egin{align}
F_{alphaeta}(mu)
= & int_0^1 frac{1}{x^alpha(1-x)^{eta}}left(sum_{n=0}^{infty}frac{(gamma)_n}{n!}mu^n x^n
ight) dx
= sum_{n=0}^{infty}frac{(gamma)_n}{n!}frac{Gamma(n+1-alpha)Gamma(1-eta)}{Gamma(n+1+gamma)}mu^n\
= & Deltasum_{n=0}^{infty}frac{(gamma)_n (1-alpha)_n}{n!(gamma+1)_n}mu^n
= Deltagamma sum_{n=0}^{infty}frac{(1-alpha)_n}{n!(gamma+n)}mu^n
end{align}$$
This implies
$$
mu^{-gamma} left(mufrac{partial}{partial mu}
ight) mu^{gamma} F_{alphaeta}(mu) =
Deltagamma sum_{n=0}^{infty}frac{(1-alpha)_n}{n!}mu^n
= Deltagammafrac{1}{(1-mu)^{1-alpha}}
$$
and hence
$$F_{alphaeta}(mu)
= Deltagamma mu^{-gamma} int_0^mu frac{
u^{gamma-1}d
u}{(1-
u)^{1-alpha}}
= Deltagamma int_0^1 frac{t^{gamma-1} dt}{(1-mu t)^{1-alpha}}
= Delta int_0^1 frac{dt}{(1 - mu t^{1/gamma})^{1-alpha}}$$
Notice if we substitute $x$ by $y = 1-x$, we have
$$F_{alphaeta}(mu) = int_0^1 frac{dy}{y^eta(1-y)^alpha(1-mu - mu y)^{gamma}}
= frac{1}{(1-mu)^gamma} F_{etaalpha}(-frac{mu}{1-mu})$$
Combine these two representations of $F_{alphaeta}(mu)$ and let $omega = left(frac{mu}{1-mu} ight)^{gamma}$, we obtain
$$F_{alphaeta}(mu) = frac{Delta}{(1-mu)^{gamma}}int_0^1 frac{dt}{( 1 + omega^{1/gamma} t^{1/gamma})^{1-eta}} = frac{Delta}{mu^gamma}int_0^omega frac{dt}{(1 + t^{1/gamma})^{1-eta}}$$
Let $(alpha,eta,gamma) = (frac14,frac12,frac14)$ and $mu = frac{sqrt{3}}{2}$, the identity we want to check becomes
$$frac{Gamma(frac34)Gamma(frac12)}{Gamma(frac54) (sqrt{3})^{1/4}}int_0^omega frac{dt}{sqrt{1+t^4}} stackrel{?}{=} frac{2sqrt{2}}{3sqrt[8]{3}} pi ag{*1}$$
Let $K(m)$ be the complete elliptic integral of the first kind associated with modulus $m$. i.e.
$$K(m) = int_0^1 frac{dx}{sqrt{(1-x^2)(1-mx^2)}}$$
It is known that $displaystyle K(frac12) = frac{8pi^{3/2}}{Gamma(-frac14)^2}$. In term of $K(frac12)$, it is easy to check $(*1)$ is equivalent to
$$int_0^omega frac{dt}{sqrt{1+t^4}} stackrel{?}{=} frac23 K(frac12) ag{*2}$$
To see whether this is the case, let $varphi(u)$ be the inverse function of above integral.
More precisely, define $varphi(u)$ by following relation:
$$u = int_0^{varphi(u)} frac{dt}{sqrt{1+t^4}}$$
Let $psi(u)$ be $frac{1}{sqrt{2}}(varphi(u) + varphi(u)^{-1})$. It is easy to check/verify
$$
varphi'(u)^2 = 1 + varphi(u)^4
implies
psi'(u)^2 = 4 (1 - psi(u)^2)(1 - frac12 psi(u)^2)
$$
Compare the ODE of $psi(u)$ with that of a Jacobi elliptic functions with modulus $m = frac12$, we find
$$psi(u) = ext{sn}(2u + ext{constant} | frac12 ) ag{*3}$$
Since we are going to deal with elliptic functions/integrals with $m = frac12$ only,
we will simplify our notations and drop all reference to modulus, i.e
$ ext{sn}(u)$ now means $ ext{sn}(u|m=frac12)$ and $K$ means $K(m = frac12)$.
Over the complex plane, it is known that $ ext{sn}(u)$ is doubly periodic with
fundamental period $4 K$ and $2i K$. It has two poles at $i K$ and $(2 + i)K$ in the fundamental domain.
When $u = 0$, we want $varphi(u) = 0$ and hence $psi(u) = infty$. So the constant
in $(*3)$ has to be one of the pole. For small and positive $u$, we want $varphi(u)$ and hence $psi(u)$ to be positive. This fixes the constant to $i K$. i.e.
$$psi(u) = ext{sn}(2u + iK )$$
and the condition $(*2)$ becomes whether following equality is true or not.
$$frac{1}{sqrt{2}} (omega + omega^{-1}) stackrel{?}{=} ext{sn}( frac43 K + i K) ag{*4}$$
Notice $ 3( frac43 K + i K) = 4 K + 3 i K $ is a pole of $ ext{sn}(u)$. if one repeat
apply the addition formula for $ ext{sn}(u+v)$
$$ ext{sn}(u+v) = frac{ ext{sn}(u) ext{cn}(v) ext{dn}(v)+ ext{sn}(v) ext{cn}(u) ext{dn}(u)}{1-m\, ext{sn}(u)^2 ext{sn}(v)^2}$$
One find in order for $ ext{sn}(3u)$ to blow up, $ ext{sn}(u)$ will be a root of
following polynomial equation:
$$3 m^2 s^8-4 m^2 s^6-4 m s^6+6 m s^4-1 = 0$$
Substitute $m = frac12$ and $s = frac{1}{sqrt{2}}(t+frac{1}{t})$ into this, the equation $omega$ need to satisfy is given by:
$$(t^8 - 6 t^4 - 3)(3 t^8 + 6 t^4 - 1 ) = 0$$
One can check that $omega = sqrt[4]{frac{sqrt{3}}{2-sqrt{3}}}$ is indeed a root of this polynomial. As a result, the original equality is valid.
Here is another approach for evaluating the integral (3).
Using **@achille hui**'s transformation, we can write
$$
I=int_0^1frac{dx}{sqrt[3]{x}sqrt{1-x}sqrt[6]{9-x}}= frac{4pi^2 2^{frac{1}{3}}}{Gammaleft(frac{1}{3}
ight)^3}int_0^frac{1}{sqrt{2}}frac{1}{sqrt{1+x^6}}dx
$$
Using the substitution $x=sqrt{t}$, we get
$$
I=frac{pi^2 2^{frac{4}{3}}}{Gammaleft(frac{1}{3}
ight)^3}int_0^frac{1}{2}frac{1}{sqrt{t+t^4}}dt
$$
We can now transform this integral into a beta integral using the substitution $t=frac{1-y}{2+y}$. This gives us
egin{align*}
I
&= frac{pi^2 2^{frac{4}{3}}}{Gammaleft(frac{1}{3}
ight)^3} int_0^1frac{1}{sqrt{1-y^3}}dy \
&= frac{pi^2 2^{frac{4}{3}}}{3Gammaleft(frac{1}{3}
ight)^3}Bleft(frac{1}{3},frac{1}{2}
ight) \
&= frac{pi^2 2^{frac{4}{3}}}{3Gammaleft(frac{1}{3}
ight)^3}left( frac{sqrt{3}Gammaleft(frac{1}{3}
ight)^3}{pi 2^{frac{4}{3}}}
ight) \
&= frac{pi}{sqrt{3}}
end{align*}
Let
$$alpha=sqrt{6} sqrt{12+7\,sqrt3}-3\,sqrt3-6\,=\,ig(2+sqrt{3}ig) ig(sqrt{2} sqrt[4]{27}-3ig)\,=\,frac{3sqrt{3}}{3+sqrt2 sqrt[4]{27}}. ag1$$
Note that $alpha$ is the unique positive root of the polynomial equation
$$alpha^4+24\,alpha^3+18\,alpha^2-27=0. ag2$$
Now consider the following integral:
$$I=int_0^1frac{dx}{sqrt[3]x sqrt[6]{1-x} sqrt{1-x\,alpha^2}}. ag3$$
I have a conjectured elementary value for it
$$Istackrel?=fracpi9Big(3+sqrt2 sqrt[4]{27}Big)=color{blue}{frac{pi}{sqrt{3}\,alpha}}. ag4$$
---
Actually, the integral $I$ can be evaluated in an exact form using _Mathematica_ or manually, using formula [DLMF 15.6.1](http://dlmf.nist.gov/15.6):
$$I=frac{4\,sqrtpi}{sqrt3}cdotfrac{Gammaleft(frac56
ight)}{Gammaleft(frac13
ight)}cdot{_2F_1}left(frac12,frac23; frac32; alpha^2
ight), ag5$$
but I could not find a way to simplify this result to $(4)$.
So, my conjecture can be restated in a different form:
$${_2F_1}left(frac12,frac23; frac32; alpha^2
ight)stackrel?=frac{sqrt2+sqrt[4]3}{4\,sqrt[4]{27}}cdotfrac{Gammaleft(frac13
ight)}{Gammaleft(frac56
ight)}cdotsqrtpi ag6$$
or
$$sum_{n=0}^inftyfrac{Gammaleft(n+frac23
ight)}{(2\,n+1)\,Gamma(n+1)}alpha^{2\,n}stackrel?=frac{3+sqrt2\,sqrt[4]{27}}{18}cdotfrac{pi^{3/2}}{Gammaleft(frac56
ight)}. ag7$$
---
The conjecture can also be given in terms of the [incomplete beta function](http://mathworld.wolfram.com/IncompleteBetaFunction.html):
$$Bleft(alpha^2; frac12,frac13
ight)stackrel?=frac{sqrtpi}2cdotfrac{Gammaleft(frac13
ight)}{Gammaleft(frac56
ight)}. ag8$$
---
> _Question:_ Is the conjecture indeed true?
_Note:_ It holds numerically up to at least $10^4$ decimal digits.
---
__Conjecture 2__
Let
$$
{small
egin{multline}
eta=frac{21}4+frac{9\,sqrt5}4-frac{15}8sqrt{750-330\,sqrt5}-frac{33}8sqrt{150-66\,sqrt5} \
+ frac12sqrt{3left(165+75\,sqrt5-46\,sqrt{750-330\,sqrt5}-103\,sqrt{150-66\,sqrt5}
ight)}.
end{multline}} ag9$$
_Added later:_ We can simplify it to
$$
smalleta=frac 3 4 left(7+3 sqrt 5-sqrt[4] 5 sqrt{66+30 sqrt 5}
ight)+frac 1 2 sqrt{495+225 sqrt 5-3 sqrt{6 ig(8545+3821 sqrt 5ig)}}. ag{$9'$}
$$
I conjecture that:
$$int_0^1frac{dx}{sqrt[3]x sqrt[6]{1-x} sqrt{1-x\,eta^2}}stackrel?=color{blue}{frac{2\,pi}{5\,sqrt3\,eta}}. ag{10}$$
I can imagine that if these conjectures are true, then there are generalizations for some other algebraic numbers.
Let $t = 1 + y^3$, we can rewrite $Bleft(alpha^2; frac12, frac13 ight)$ as
$$int_0^{alpha^2} t^{-1/2} (1-t)^{-2/3} dt
= int_{-1}^{-sqrt[3]{1-alpha^2}} frac{3y^2dy}{sqrt{1+y^3} y^2}
= 3 int_{-1}^{-sqrt[3]{1-alpha^2}} frac{dy}{sqrt{1+y^3}}$$
Following the setup in my [answer](https://math.stackexchange.com/a/692107/59379) to a related question. Let
$;displaystyleeta = frac{Gammaleft(frac13
ight)Gammaleft(frac16
ight)}{sqrt{3pi}};$ and $wp(z)$ be the [Weierstrass elliptic $wp$ function](http://en.wikipedia.org/wiki/Weierstrass%27s_elliptic_function) with fundamental periods $1$ and $e^{ipi/3}$. $wp(z)$ is known to satisfy an ODE of the form
$$wp'(z)^2 = 4 wp(z)^3 - g_2 wp(z) - g_3quad ext{ where }quad g_2 = 0 ; ext{ and };g_3 = frac{eta^6}{16}$$
If one perform variable substitution $;displaystyle y = -frac{4}{eta^2} wpleft(frac{iz}{eta}
ight)$, one has
$$frac{dy}{sqrt{1+y^3}} = -dzquad ext{ and }quad
egin{cases}
yleft(frac{sqrt{3}eta}{3}
ight)
= -frac{4}{eta^2}wpleft(ifrac{sqrt{3}}{3}
ight) = 0\
\
yleft(frac{sqrt{3}eta}{2}
ight)
= -frac{4}{eta^2}wpleft(ifrac{sqrt{3}}{2}
ight) = -1
end{cases}$$
Using this, we can express conjecture $(8)$ in terms of $y(cdot)$ and/or $wp(cdot)$:
$$egin{align}
& Bleft(alpha^2; frac12, frac13
ight)
stackrel{?}{=} frac{sqrt{pi}}{2}frac{Gammaleft(frac13
ight)}{Gammaleft(frac56
ight)} = frac{sqrt{3}}{4}eta\
iff & 3left[y^{-1}(-1) - y^{-1}(-sqrt[3]{1-alpha^2})
ight]
stackrel{?}{=} frac{sqrt{3}}{4}eta\
iff & y^{-1}(-sqrt[3]{1-alpha^2})
stackrel{?}{=} frac{5sqrt{3}}{12}eta\
iff & frac{4}{eta^2}wpleft(ifrac{5sqrt{3}}{12}
ight)
stackrel{?}{=} sqrt[3]{1-alpha^2}
end{align}
$$
Let $u_0 = ifrac{sqrt{3}}{3}$, $u_{-1} = ifrac{sqrt{3}}{2}$ and $u = ifrac{5sqrt{3}}{12} = frac12(u_0 + u_{-1})$. Using the addition formula for $wp$ function,
we have
$$
wp(2u) = wp(u_0 + u_{-1})
= frac14left[frac{wp'(u_0)-wp'(u_{-1})}{wp(u_0)-wp(u_{-1})}
ight]^2 - wp(u_0) - wp(u_{-1})\
=frac14left[frac{-ifrac{eta^3}{4} - 0}{0 - frac{eta^2}{4}}
ight]^2 - 0 - frac{eta^2}{4}
= -frac{eta^2}{2}
$$
Using the duplication formula of $wp$ function, we get
$$
-frac{eta^2}{2} = wp(2u) = frac14left(frac{(6wp(u)^2-frac12 g_2)^2}{4wp(u)^3-g_2wp(u)-g_3}
ight) -2wp(u)
= frac{9wp(u)^4}{4wp(u)^3-frac{eta^4}{16}} - 2wp(u)
$$
Let $Y = frac{4}{eta^2}wp(u)$ and $A^2 = 1 - Y^3$, above condition is equivalent to
$$egin{align}
& Y^4 + 8 Y^3 + 8 Y - 8 = 0 ag{*1a}\
iff & Y(8+Y^3) = 8(1-Y^3) ag{*1b}\
implies & (9-A^2)^3(1-A^2) = 512A^6 ag{*1c}\
iff & (A^4-24A^3+18A^2-27)(A^4+24A^3+18A^2-27) = 0 ag{*1d}
end{align}$$
+ Since $alpha$ is a root for one of the factors in $(*1d)$, $A = alpha$ satisfies $(*1d)$ and hence $(*1c)$.
+ Since $0 < alpha < 1$ implies $1 - alpha^2 > 0$, $(*1c) implies (*1b)$ in this particular case.
i.e. $Y = sqrt[3]{1-alpha^2}$ satisfies $(*1b)$ and hence $(*1a)$.
+ Since $u$ lies between $u_0$ and $u_{-1}$, $frac{4}{eta^2}wp(u) > 0$. Using
the fact $(*1a)$ has only one positive root, we find $frac{4}{eta^2}wp(u) = sqrt[3]{1-alpha^2}$. i.e. conjecture $(8)$ is true.
$defBeta{B}def frac#1#2{{ extstylefrac{#1}{#2}}}$
Perhaps this might be helpful to someone. The integral is equal to, as you note,
$$ J(y) = int_0^1 x^{-1/3}(1-x)^{-1/6}(1-xy^2)^{-1/2}\,dx =
frac{2pi}{ysqrt{3}} frac{Beta(y^2,frac12,frac13)}{Beta(frac12,frac13)}, $$
where $Beta(z,a,b)$ is the incomplete beta function.
Consider the function
$$ I(y^2,a,b) = frac{Beta(y^2,a,b)}{Beta(a,b)}, $$
and rewrite using [DLMF 8.17](http://dlmf.nist.gov/8.17) it as
$$ I(y^2, frac12, frac13) = 1-2I(z, frac13, frac13), qquad 4z(1-z) = 1-y^2. $$
Then the function
$$ f(z) = I(z,{ extstylefrac13,frac13}) = 3z^{1/3}frac{{}_2F_1( frac13, frac23; frac43;z)}{Beta(frac13,frac13)} $$
is the one for which the conjectures are equivalent to (choosing the right root $z$):
$$ f(z) = frac14 quadLeftrightarrowquad z^4-14z^3+24z^2-14z+1=0, $$
$$ f(z) = frac25 quadLeftrightarrowquad 1+17 z-107 z^2+164 z^3-155 z^4+164 z^5-107 z^6+17 z^7+z^8=0, $$
From my tests it appears (I don't know how to prove this)
that the function $z(w)$ which solves
the equation $f(z(w)) = w$ always has algebraic values when $w$
is a rational number. The original integral is
$frac{2pi}{ysqrt{3}}(1-2f(z))$, which is algebraic times $pi$ when $f(z)$
is rational and $z$ algebraic, so I think the question is really about solving $f(z)=w$.
How to prove the following conjectured identity?
$$int_0^inftyfrac{dx}{sqrt[4]{7+cosh x}}stackrel{color{#a0a0a0}?}=frac{sqrt[4]6}{3sqrtpi}Gamma^2ig( frac14ig) ag1$$
It holds numerically with precision of at least $1000$ decimal digits.
Are there any other integers under the radical except $7$ and $1$ that result in a nice closed form?
I will follow @user15302's idea. In [**this answer**](https://math.stackexchange.com/questions/1379472/closed-form-of-int-0-infty-frac1-lefta-cosh-x-right1-n-dx-for/1379605#1379605), I showed that
$$ int_{0}^{infty} frac{dx}{(a + cosh x)^{s}} \, dx = frac{1}{(a+1)^{s}} int_{0}^{1} frac{v^{s-1}}{sqrt{(1-v)(1-bv)}} \, dv, $$
where $b = frac{a-1}{a+1}$. Now let $I$ denote the Vladimir's integral and set $s = frac{1}{4}$ and $a = 7$. Then we have $b = frac{3}{4}$ and
$$ I = 2^{-3/4} int_{0}^{1} frac{1}{v^{3/4}sqrt{(1-v)(1-frac{3}{4}v)}} \, dv. $$
The reason why the case $b = frac{3}{4}$ is special is that, if we plug $v = operatorname{sech}^2 t$ then we can utilize the triple angle formula to get the following surprisingly neat integral
$$ I = 2^{5/4} int_{0}^{infty} frac{cosh t}{sqrt{cosh 3t}} \, dt. $$
Now using the substitution $x = e^{-6t}$, we easily find that
$$ I = frac{1}{3 sqrt[4]{2}} int_{0}^{1} frac{x^{-11/12} + u^{-7/12}}{sqrt{1+x}} \, dx = frac{1}{3 sqrt[4]{2}} int_{0}^{infty} frac{dx}{x^{11/12}sqrt{1+x}}. $$
The last integral can be easily calculated by the following formula
$$ int_{0}^{infty} frac{x^{a-1}}{(1+x)^{a+b}} \, dx = eta(a, b) = frac{Gamma(a)Gamma(b)}{Gamma(a+b)}. $$
Therefore we obtain the following closed form
$$ I = frac{Gamma(frac{1}{12})Gamma(frac{5}{12})}{3 sqrt[4]{2}sqrt{pi}}. $$
In order to verify that this is exactly the same as Vladimir's result, We utilize the [*Legendre multiplication formula*](https://en.wikipedia.org/wiki/Multiplication_theorem#Gamma_function-Legendre_function) and the [*reflection formula*](https://en.wikipedia.org/wiki/Reflection_formula) to find that
$$ Gamma( frac{1}{12})Gamma( frac{5}{12})
= frac{Gamma( frac{1}{12})Gamma( frac{5}{12})Gamma( frac{9}{12})}{Gamma( frac{3}{4})}
= frac{2 pi cdot 3^{1/4} Gamma(frac{1}{4})}{Gamma( frac{3}{4})}
= 2^{1/2} 3^{1/4} Gamma( frac{1}{4})^2. $$
This completes the proof.
By replacing $x$ with $4u$, then $cosh u$ with $frac{1}{t}$, we have:
$$ I = frac{1}{2^{3/4}}int_{0}^{1}frac{dt}{(1-t^2+t^4)^{1/4}(1-t^2)^{1/2}}=frac{1}{2^{3/4}}int_{0}^{1/2}frac{dt}{(1-t(1-t))^{1/4}(t(1-t))^{1/2}} $$
Next, by replacing $t(1-t)$ with $v/4$,
$$ I=frac{1}{2^{11/4}}int_{0}^{1}frac{dv}{(1-v/4)^{1/4}(v(1-v))^{1/2}}$$
then, by setting $v=4-3z$,
$$ I = frac{3^{1/4}}{2^{9/4}}int_{1}^{4/3}frac{dz}{z^{1/4}((4-3z)(1+z))^{1/2}}=frac{3^{1/4}}{2^{5/4}}int_{1}^{2/sqrt{3}}sqrt{frac{z}{(4-3z^2)(1+z^2)}}\,dz$$
that, at least, looks manageable. We also have:
$$ I = frac{1}{2^{1/4}}int_{0}^{1}frac{du}{(3u^4+u^2)^{1/4}(1-u^2)^{1/2}} ag{1}$$
that Mathematica gladly evaluates to:
$$ I = frac{2^{1/4}\,Gammaleft(frac{1}{4}
ight)^2}{3^{3/4}sqrt{pi}}. $$
Now we just need to understand *how*.
----------
I think this problem can be solved by invoking the theory of $j$-invariants for (hyper?)-elliptic curves, but I am not so confident in the topic to find the right change of variables that brings our integral into a complete elliptic integral. I think that [Noam Elkies][1] would solve this problem in a few seconds, so I am asking his help.
----------
**Update**. Found. Our claim was proven by Zucker and Joyce in [Special values of the hypergeometric series II][2], it is the result $(7!cdot! 6)$. It is derived through standard hypergeometric manipulations, by starting with the elliptic modulus $k$ for which:
$$frac{K'(k)}{K(k)}=3.$$
The modular function to be considered for regarding our integral as a period is so the [elliptic lambda function][3].
[1]: https://math.stackexchange.com/users/93983/noam-d-elkies
[2]: http://www.researchgate.net/publication/231927689_Special_values_of_the_hypergeometric_series_III
[3]: http://mathworld.wolfram.com/EllipticLambdaFunction.html
Take the integral in the form $$I=frac{1}{2^{1/4}}int_{0}^{1}frac{du}{left(3u^{4}+u^{2}
ight)^{1/4}left(1-u^{2}
ight)^{1/2}}=frac{1}{2^{1/4}}int_{0}^{1}frac{du}{left(3u^{2}+1
ight)^{1/4}left(1-u^{2}
ight)^{1/2}left(u^{2}
ight)^{1/4}}
$$ then put $u^{2}=s
$ $$=frac{1}{2^{5/4}}int_{0}^{1}frac{ds}{left(3s+1
ight)^{1/4}left(1-s
ight)^{1/2}s^{3/4}}
$$ and now put $s=1-t
$ $$=frac{1}{2^{7/4}}int_{0}^{1}frac{dt}{left(1-3t/4
ight)^{1/4}left(1-t
ight)^{3/4}t^{1/2}}.
$$ Now recalling the identity $$\,_{2}F_{1}left(a,b;c;z
ight)=frac{Gammaleft(c
ight)}{Gammaleft(b
ight)Gammaleft(c-b
ight)}int_{0}^{1}frac{t^{b-1}left(1-t
ight)^{c-b-1}}{left(1-tz
ight)^{a}}dt
$$ we have $$I=frac{1}{2^{7/4}}frac{Gammaleft(frac{1}{2}
ight)Gammaleft(frac{1}{4}
ight)}{Gammaleft(frac{3}{4}
ight)}\,_{2}F_{1}left(frac{1}{4},frac{1}{2};frac{3}{4};frac{3}{4}
ight)
$$ and in this case it is possible calculate the exact value of the hypergeometric function (see the update in the Jack D'Aurizio's answer for reference) $$\,_{2}F_{1}left(frac{1}{4},frac{1}{2};frac{3}{4};frac{3}{4}
ight)=frac{2sqrt{2}}{3^{3/4}}
$$ and so $$I=frac{Gamma^{2}left(frac{1}{4}
ight)}{6^{3/4}sqrt{pi}}.
$$ The result is not equal to $sqrt[4]{6}Gamma^{2}left(frac{1}{4}
ight)/left(2sqrt{pi}
ight)
$ but I haven't found a mistake in my calculations.
It is quite a well known fact that:
$$int_0^{+infty} frac{sin x}{x} \, dx = frac{pi}{2}$$
also the value of related series is very similiar:
$$sum_{n = 1}^{+infty} frac{sin n}{n} = frac{pi - 1}{2}$$
Combining these two identities and using ${
m sinc}$ function we get:
$$int_{-infty}^{+infty} {
m sinc}\, x \, dx = sum_{n = -infty}^{+infty} {
m sinc}\, n = pi$$
What is more interesting is the fact that the equality:
$$int_{-infty}^{+infty} {
m sinc}^k\, x \, dx = sum_{n = -infty}^{+infty} {
m sinc}^k\, n$$
holds for $k = 1,2,ldots, 6$. There are some other nice identities with ${
m sinc}$ where sum equals integral but moving on to other functions we have e.g.:
$$sum_{n = -infty}^{+infty} inom{alpha}{n} e^{int} = int_{-infty}^{+infty} inom{alpha}{n} e^{itx} \, dx = (1+e^{it})^alpha, ; alpha >-1$$which is due to Pollard & Shisha.
And finally the identity which is related to the famous *Sophomore's Dream*:
$$int_0^1 frac{dx}{x^x} = sum_{n = 1}^{+infty} frac{1}{n^n}$$
Unfortunately in this case the summation range is not even close to the interval of integration.
Do you know any other interesting identities which show that "sum = integral"?
Several papers are dedicated to the subject of integrals of functions that equal the sum of the same function, primarily for estimation purposes.
[Boas and Pollard][1] (1973) has some interesting sum-integral equalities:
$$pi/alpha=sum_{n=-infty}^infty frac{sin^2 (c+n)alpha}{(c+n)^2}=int_{-infty}^infty frac{sin^2 (c+n)alpha}{(c+n)^2}\, ext{d}n$$
$$pioperatorname{sgn} a=sum_{n=-infty}^infty frac{sin (n+c)alpha}{n+c}=int_{-infty}^infty frac{sin (n+c)alpha}{n+c}\, ext{d}n$$
It also gives several general formulae for functions that suffice:
$$sum_{n=-infty}^infty f(n)=int_{-infty}^infty f(n) \, ext{d}n$$
mainly with Fourier analysis.
---
[This][2] paper gives an equality with the [Bessel J function][3]:
$$int_{-infty}^infty frac{J_y (at) J_y(bt)}{t}\, ext{d}t=sum_{t=-infty}^infty frac{J_y (at) J_y(bt)}{t}$$
and some more references:
> There have been a number of studies of this kind of sum-integral
> equality by various groups, for example, Krishnan & Bhatia in the
> 1940s (Bhatia & Krishnan 1948; Krishnan 1948a,b; Simon 2002) and Boas,
> Pollard & Shisha in the 1970s (Boas & Stutz 1971; Pollard & Shisha
> 1972; Boas & Pollard 1973).
---
See also [Surprising Sinc Sums and Integrals][4] which has some other equalities. This paper also states that (paraphrasing)
*If $G$ is of bounded variation on $[−delta, delta]$, vanishes outside $(−α, α)$, is Lebesgue integrable over $(−α, α)$ with $0 < α < 2pi$ and has a Fourier transform of $g$, then*
$$sum_{n=infty}^infty g(n)=int_{-infty}^infty g(x)\, ext{d}x+sqrt{frac{pi}{2}}(G(0-)+G(0+))$$
---
Ramanujan's [second lost notebook][5] contains some sums of functions that equal the integral of their functions (Chapter 14, entries 5(i), 5(ii), 16(i), 16(ii)).
---
If you want, even more references with examples are in the papers I have mentioned.
[1]: http://carma.newcastle.edu.au/jon/Preprints/Papers/Published-InPress/Sinc-sums/Related/boas.pdf
[2]: http://rsc.anu.edu.au/~pgill/papers/159_Bessel.pdf
[3]: http://en.wikipedia.org/wiki/Bessel_function#Bessel_functions_of_the_first_kind_:_J.CE.B1
[4]: http://web.cs.dal.ca/~jborwein/sinc-sums.pdf
[5]: http://www.scribd.com/doc/48940173/Ramanujan-Notebooks-II-pdf
How one can prove that the infinite sum of this function equals its integral
$$
sum_{n=-infty}^inftyfrac{cospisqrt{n^2+1}}{3+4n^2}=int_{-infty}^inftyfrac{cospisqrt{x^2+1}}{3+4x^2}dx ? ag{1}
$$
My analysis: Mathematica wasn't able to return any closed form for the integral or the sum. Then I checked this relation by numerical computations and it agreed to about 20 decimal places.
I know from this question [Sum equals integral][1] that the function $ ext{sinc} x=frac{sin x}{x}$ has the same property
$$
int_{-infty}^{+infty} {
m sinc}\, x \, dx = sum_{n = -infty}^{+infty} {
m sinc}\, n = pi
$$
I tried to find a closed form for the integral $(1)$ but couldn't.
Motivation: I was challenged by a friend to prove this relation. I'm curious how one can prove it?
Note: There has been a suggestion to straightforwardly apply Euler-MacLauren summation formula to prove this statement. Though I don't know why it can not be applied in this case, I checked numerically whether the sum equals the integral for the similar looking functions $f_1(x)=frac{cospisqrt{x^2+1}}{1+x^2}$ and $f_2(x)=frac{cospisqrt{x^2+1}}{2+x^2}$, but in both cases there was a difference of about 1% between the sum and the integral. In starck contrast to this, using the same algorithm for $frac{cospisqrt{x^2+1}}{3+4x^2}$ there wasn't any difference between the sum and the integral at least to 20 decimal places. So I think it is very unlikely that 1% error can be attributed to computational error.
[1]: https://math.stackexchange.com/questions/170747/sum-equals-integral
For any positive $a$, define
$$f_a(x) =frac{cospisqrt{x^2+1}}{x^2+a^2}$$
What you have observed is caused by the equality
$$sum_{n=-infty}^infty f_a(n) - int_{-infty}^infty f_a(x) dx =
frac{2pi}{a(e^{2pi a} - 1)} imes
egin{cases}
coshpisqrt{a^2-1}, & a > 1\
\
cospisqrt{1-a^2}, & a < 1
end{cases}
ag{*1}
$$
and the fact $$cospisqrt{1-a^2} = cosfrac{pi}{2} = 0quad ext{ when } a^2 = frac34$$
To see why $(*1)$ is true, we use the fact $f_a(n)$ is an even function in $n$ to rewrite LHS of $(*1)$ as
$$2 left[sum_{n=0}^infty f_a(n) - left(int_0^infty f_a(x) dx + frac12 f_a(0) ight) ight]$$
This is similar to what you will find in
the [Abel-Plana formula](https://en.wikipedia.org/wiki/Abel%E2%80%93Plana_formula)${}^{color{blue}{[1]}}$,
> For any function $f(z)$ which is
>
> 1. continuous on $Re z ge 0$ and analytic on $Re z > 0$
> 2. $f(z) sim o(e^{2pi|Im z|} )$ as $Im z o pm infty$, uniformly with respect to $Re z$.
> 3. $f(z) sim O(e^{2pi|Im z|}/|z|^{1+epsilon})$ as $Re z o +infty$ ${}^{color{blue}{[2]}}$.
>
> we have
>
$$sum_{n=0}^infty f(n) = int_0^infty f(x) dx + frac12 f(0) + i int_0^infty frac{f(it) - f(-it)}{e^{2pi t}-1} dt ag{*2}$$
However, $f_a(x)$ doesn't exactly satisfy the condition above. It has two
poles at $pm a i$. After a little bit of tweaking of the contour used in the proof of the Abel-Plana formula, one find:
$$ ext{LHS}(*1) = 2i lim_{epsilon o 0^{+}} int_0^infty frac{f_a(it+epsilon) - f_a(-it+epsilon)}{e^{2pi t} - 1} dt$$
For $t
e a$, since $f_a(z)$ is even, the two pieces in $f_a(it+epsilon) - f_a(-it+epsilon)$ cancels out as $epsilon o 0^{+}$.
For $t approx a$, the two pieces can be combined to a integral of $frac{f(it)}{e^{2pi t}-1}$ over a circle centered at $a$.
As a result, RHS reduces to
$$(2i)(2pi i) ext{Res}_{t = a}left[frac{cospisqrt{1-t^2}}{(a^2 - t^2)(e^{2pi t} - 1)}
ight]
= frac{2pi}{a(e^{2pi a}-1)} imesegin{cases}
coshpisqrt{a^2-1}, & a > 1\
\
cospisqrt{1-a^2}, & a < 1
end{cases}
$$
Back to the special case $a^2 = frac{3}{4}$ which corresponds to the equality in question.
When $a^2 = frac{3}{4}$, the "pole" of $f_a(z)$ at $z = pm a i$ become removable singularities. The original version of Abel-Plana formula in $(*2)$ applies. Since $f_a(x)$ is even, last integral in $(*2)$ vanishes and the equality follows. This explain why the sum equal to the integral for $frac{cospisqrt{x^2+1}}{3+4x^2}$ but not other similar looking integrand like $frac{cospisqrt{x^2+1}}{1+x^2}$ or $frac{cospisqrt{x^2+1}}{2+x^2}$.
**Notes**
+ $color{blue}{[1]}$ For more details of Abel-Plana formula and its derivation, please refer to $S 8.3$ of Frank W. J Olver's book: [Asymptotics and Special Functions](http://www.amazon.com/Asymptotics-Special-Functions-Computer-mathematics/dp/012525850X).
+ $color{blue}{[2]}$ In order to convert the AP formula on finite sum in Olver's book to infinite sum here, I have added a condition $(3)$ for this particular problem. The whole purpose of that is to force following limits to zero.
$$limlimits_{b oinfty} f(b) = 0quad ext{ and }quadlimlimits_{b oinfty}int_0^infty frac{f(b+it)-f(b-it)}{e^{2pi t} - 1}dt = 0$$
For other $f(z)$, if one can justifies these limits, we can forget condition $(3)$ and the AP formula remains valid.