[Largeint_{0}^{1}frac{arctan x}{sqrt{1-x^{2}}}mathrm{d}x
]
(Largemathbf{Solution:})
首先第一种做法，含参积分.不多说直接上图
第二种方法则是利用级数，易知
[egin{align*}
int_{0}^{1}frac{arctan x}{sqrt{1-x^{2}}}mathrm{d}x&=int_0^{pi/2}arctan(sin(x))\,mathrm{d}x\&=sum_{k=0}^inftyfrac{(-1)^k}{2k+1}int_0^{pi/2}sin^{2k+1}(x)\,mathrm{d}x\
&=sum_{k=0}^inftyfrac{(-1)^k}{2k+1}frac{2^k\,k!}{(2k+1)!!}\
&=sum_{k=0}^inftyfrac{(-1)^k}{(2k+1)^2}frac{4^k}{displaystyleinom{2k}{k}}
end{align*}]
下面来解决最后一个级数，利用Beta函数我们可以得到以下等式
[frac1{displaystyleinom{2n}{n}}=(2n+1)int_0^1t^n(1-t)^nmathrm{d}t
]
所以
[egin{align*}
sum_{n=0}^inftyfrac{(-4)^nx^{2n}}{(2n+1)displaystyleinom{2n}{n}}
&=int_0^1frac1{1+4x^2t(1-t)}mathrm{d}t\
&=int_0^1frac1{1+x^2-x^2(2t-1)^2}mathrm{d}t\
&=frac1{1+x^2}int_0^1frac1{1-dfrac{x^2}{1+x^2}(2t-1)^2}mathrm{d}t\
&=frac1{1+x^2}int_{-1}^1frac1{1-dfrac{x^2}{1+x^2}t^2}frac12mathrm{d}t\
&=frac1{2xsqrt{1+x^2}}int_{-x/sqrt{1+x^2}}^{x/sqrt{1+x^2}}frac1{1-t^2}mathrm{d}t\
&=frac1{xsqrt{1+x^2}}mathrm{arctanh}left(frac{x}{sqrt{1+x^2}}
ight)\
&=frac1{xsqrt{1+x^2}}mathrm{arcsinh}(x)
end{align*}]
两边积分可以得到
[egin{align*}
sum_{n=0}^inftyfrac{(-4)^n}{(2n+1)^2displaystyle inom{2n}{n}}
&=int_0^1frac1{xsqrt{1+x^2}}mathrm{arcsinh}(x)\,mathrm{d}x\
&=-int_0^1mathrm{arcsinh}(x)frac1{sqrt{vphantom{ig|}1+1/x^2}}mathrm{d}frac1x\
&=-int_0^1mathrm{arcsinh}(x)\,mathrm{d}\,mathrm{arcsinh}left(frac1x
ight)\
&=-\,mathrm{arcsinh}^2(1)+int_0^1mathrm{arcsinh}left(frac1x
ight)\,mathrm{d}\,mathrm{arcsinh}(x)\
&=-\,mathrm{arcsinh}^2(1)-int_1^inftymathrm{arcsinh}(x)\,mathrm{d}\,mathrm{arcsinh}left(frac1x
ight)\
&=-\,mathrm{arcsinh}^2(1)+int_1^inftyfrac1{xsqrt{1+x^2}}mathrm{arcsinh}(x)\,mathrm{d}x\
&=-frac12\,mathrm{arcsinh}^2(1)+frac12int_0^inftyfrac1{xsqrt{1+x^2}}mathrm{arcsinh}(x)\,mathrm{d}x\
&=-frac12\,mathrm{arcsinh}^2(1)+frac12int_0^inftyfrac{t\,mathrm{d}t}{sinh(t)}
end{align*}]
其中
[int_0^inftyfrac{t\,mathrm{dt}}{sinh(t)}=int_0^inftysum_{k=0}^infty2t\,e^{-(2k+1)t}\,mathrm{d}t=sum_{k=0}^inftyfrac2{(2k+1)^2}=frac{pi^2}4
]
所以
[color{red}{sum_{n=0}^inftyfrac{(-4)^n}{(2n+1)^2displaystyle inom{2n}{n}}=frac{pi^2}8-frac12mathrm{arcsinh}^2(1)}
]
即
[Largeoxed{displaystyle int_{0}^{1}frac{arctan x}{sqrt{1-x^{2}}}\, mathrm{d}x=color{blue}{frac{pi^2}8-frac12mathrm{arcsinh}^2(1)}}
]