[Largedisplaystyle int_0^infty frac{ln left(1+dfrac{pi^2}{4x}
ight)}{e^{sqrt{x}}-1}mathrm{d}x
]
(Largemathbf{Solution:})
Step 1 - Split
Let (displaystyle I=int_0^infty frac{ln left(1+dfrac{pi^2}{4x}
ight)}{e^{sqrt{x}}-1}mathrm{d}x). Now put (xmapsto x^2)
[egin{align*} I &=2int_0^infty frac{xlnleft( 1+dfrac{pi^2}{4x^2}
ight)}{e^x-1}mathrm{d}x\&= 2int_0^infty frac{x ln left( 4x^2+pi^2
ight)-xln (4)-2xln(x)}{e^x-1}mathrm{d}x \ &= 2underbrace{int_0^infty frac{x ln left( 4x^2+pi^2
ight)}{e^x-1}mathrm{d}x}_{=I_1}-4ln(2) underbrace{int_0^infty frac{x}{e^x-1}mathrm{d}x}_{=I_2}-4 underbrace{int_0^infty frac{xln(x)}{e^x-1}mathrm{d}x}_{=I_3} ag{1} end{align*}
]
Step 2 - Evaluation of (I_3)
Note that for (Re (z)>1), we have
[egin{align*} int_0^infty frac{x^{z-1}}{e^x-1}mathrm{d}x = Gamma(z)zeta(z) ag{2}end{align*}
]
Differentiate both sides with respect to (z).
[egin{align*} int_0^infty frac{x^{z-1}ln(x)}{e^x-1}mathrm{d}x= psi_0(z)Gamma(z) zeta(z)+Gamma(z)zeta '(z) ag{3}end{align*}
]
Put (z=2) to obtain
[egin{align*}int_0^infty frac{x ln(x)}{e^x-1}mathrm{d}x &= frac{pi^2}{6}(1-gamma)-frac{pi^2}{6} left(12 lnmathbf{A}-gamma-ln(2pi)
ight) \&= frac{pi^2}{6} left( 1+ln(2pi)-12 lnmathbf{A}
ight) ag{4}end{align*}
]
Step 3 - Evaluation of (I_2)
(I_2) can be evaluated easily by means of equation (2).
[egin{align*}
int_0^infty frac{x}{e^x-1}mathrm{d}x = frac{pi^2}{6} ag{5}
end{align*}]
Step 4 - Evaluation of (I_1)
From here begins the dirty job. We require equation (4) of Adamchik's paper.
[egin{align*} int_0^infty frac{x ln(x^2+z^2)}{e^{2pi x}-1}mathrm{d}x &= zeta'(-1,z)-frac{z^2}{2}ln z+frac{z^2}{4}+frac{z}{2}ln z \
&~~~-2z int_0^infty frac{arctan left( dfrac{x}{z}
ight)}{e^{2pi x}-1}mathrm{d}x quad Re(z)>0 end{align*}]
Now, from Binet's second formula we have
[egin{align*} int_0^infty frac{x ln(x^2+z^2)}{e^{2pi x}-1}mathrm{d}x &= zeta'(-1,z)-frac{z^2}{2}ln z+frac{z^2}{4}+frac{z}{2}ln z \ &quad -z left{ln Gamma(z) -left(z-frac{1}{2}
ight)ln z+z -frac{1}{2}ln(2pi)
ight} ag{6} end{align*}
]
Set (x mapsto dfrac{x}{2pi}) and put (z=dfrac{1}{4}).
[egin{align*} &frac{1}{4pi^2}int_0^infty frac{x ln left( 4x^2+pi^2
ight)}{e^x-1}mathrm{d}x-frac{ln 2+ln(2pi)}{12} \=& zeta' left(-1, frac{1}{4}
ight)+frac{1}{16}ln(2)+frac{1}{64}-frac{1}{4}ln(2)-frac{1}{4} left{ln Gamma left(frac{1}{4}
ight) -frac{ln (2)}{2}+frac{1}{4} -frac{1}{2}ln(2pi)
ight} \ &int_0^infty frac{x ln left(4x^2 +pi^2
ight)}{e^x-1}mathrm{d}x \=& 4pi^2 zeta' left( -1,frac{1}{4}
ight)+frac{pi^2}{12}ln(2)+frac{5}{6}pi^2 ln(2pi) -frac{3}{16}pi^2-pi^2 ln Gamma left( frac{1}{4}
ight) ag{7} end{align*}
]
Now, from equations (3) and (11) of Adamchik's paper we get
[egin{align*} zeta' left(-1,frac{1}{4}
ight) &= zeta'(-1)-frac{3}{4}ln Gamma left( frac{1}{4}
ight)-ln mathrm{G} left( frac{1}{4}
ight) \ &= frac{1}{12}-lnmathbf{A}-frac{3}{4}ln Gamma left( frac{1}{4}
ight)-left(frac{3}{32} -frac{mathbf{G}}{4pi}-frac{3}{4}ln Gamma left( frac{1}{4}
ight)-frac{9}{8}lnmathbf{A}
ight) \ &= frac{lnmathbf{A}}{8}+frac{mathbf{G}}{4pi}-frac{1}{96} ag{8} end{align*}
]
where (mathrm{G}(z)) denotes the Barnes G Function.
Using this result in (7), we get
[egin{align*} &int_0^infty frac{x ln left(4x^2 +pi^2
ight)}{e^x-1}mathrm{d}x \&= 4pi^2 left( frac{lnmathbf{A}}{8}+frac{mathbf{G}}{4pi}-frac{1}{96}
ight)+frac{pi^2}{12}ln(2)+frac{5}{6}pi^2 ln(2pi) -frac{3}{16}pi^2-pi^2 ln Gamma left( frac{1}{4}
ight) \ &= frac{pi^2}{2}lnmathbf{A}+frac{pi^2}{12}ln(2)-frac{11}{48}pi^2+frac{5pi^2}{6}ln(2pi) +pi mathbf{G} - pi^2 ln Gamma left( frac{1}{4}
ight) ag{9} end{align*}
]
Step 5 - Final Answer
Combining everything with the help of equation (1), we get
[egin{align*} I &= 2 left(frac{pi^2}{2}lnmathbf{A}+frac{pi^2}{12}ln(2)-frac{11}{48}pi^2+frac{5pi^2}{6}ln(2pi) +pi mathbf{G} - pi^2 ln Gamma left( frac{1}{4}
ight)
ight) \&~~~-4ln(2) left(frac{pi^2}{6}
ight)-4 left( frac{pi^2}{6} left( 1+ln(2pi)-12 lnmathbf{A}
ight)
ight) \ &=Largeoxed{displaystylecolor{blue}{pi^2 left{ ln left( frac{pi mathbf{A}^9 sqrt{2}}{Gamma^2 left( dfrac{1}{4}
ight)}
ight)-frac{9}{8}
ight}+2pi mathbf{G}}} end{align*}
]