I am trying to evaluate the following integral, so far I tried 2 different ways, but could not finish the proof. Through the second method, it seems that I got closer
$$\int_{0}^{\pi/2}\arctan(\sin(x))dx=\frac{\pi^2}{8}-\frac{\ln^2(\sqrt{2}-1)}{2}$$
First Method
Consider the more general version with parameter k
$$I(k)=\int_{0}^{\pi/2}\arctan(k\sin(x))dx$$
$$I^{\prime}(k)=\int_{0}^{\pi/2}\frac{\sin(x)}{1+k^2\sin^2(x)}dx$$
$$I^{\prime}(k)=\int_{0}^{\pi/2}\frac{\sin(x)}{\cos^2(x)+\sin^2(x)+k^2\sin^2(x)}dx$$
$$I^{\prime}(k)=\int_{0}^{\pi/2}\frac{\sin(x)}{\cos^2(x)+(1+k^2)\sin^2(x)}\frac{1}{\frac{\sin^2(x)}{\sin^2(x)}}dx$$
$$I^{\prime}(k)=\int_{0}^{\pi/2}\frac{\csc(x)}{\cot^2(x)+(1+k^2)}dx$$
substitution $\cot^2(x)=t$ does not seem very helpful.
Second Method
let $\sin(x)\longrightarrow x$
$$\int_{0}^{\pi/2}\arctan(\sin(x))dx=\int_{0}^{1}\frac{\arctan(x)}{\sqrt{1-x^2}}dx$$
integrating by parts
$$\int_{0}^{1}\frac{\arctan(x)}{\sqrt{1-x^2}}dx=\arctan(x)\cdot\arcsin(x)|_{0}^{1}-\int_{0}^{1}\frac{\arcsin(x)}{1+x^2}dx$$
$$=\frac{\pi^2}{8}-\underbrace{\int_{0}^{1}\frac{\arcsin(x)}{1+x^2}dx}_{J}$$
using the expansion of $\arcsin(x)$
$$J=\int_{0}^{1}\frac{\arcsin(x)}{1+x^2}dx=\sum_{n=0}^{\infty}\frac{(2n)!}{2^{2n}(n!)^2}\frac{1}{2n+1}\int_{0}^{1}\frac{x^{2n+1}}{1+x^2}dx$$
Can someone indicate a method to help me finish at least one of the two methods?
Thank you
Best Answer
First method:
$$I(k)=\int_{0}^{\pi/2}\arctan(k\sin x)dx,\>\>\>\>\> I^{\prime}(k)=\int_{0}^{\pi/2}\frac{\sin x}{1+k^2\sin^2x}dx$$ Evaluate $I’(k)$ with $t=\cos x$ \begin{align} I^{\prime}(k)=\int_{0}^{1}\frac{1}{1+k^2-k^2 t^2}dt =\frac{\text{arctanh}\frac{k}{\sqrt{1+k^2}}}{k\sqrt{1+k^2}} = \frac{\text{arcsinh}{\>k}}{k\sqrt{1+k^2}} \end{align} Then, $I(0)=0$, $I(\infty) =\frac{\pi^2}4 = \int_0^\infty I’(k)dk$
\begin{align} &\int_{0}^{\pi/2}\arctan(\sin x)dx =I(1)= \int_0^1 I’(k)dk \\ =&\int_{0}^{1}\frac{\text{arcsinh}{\>k}}{k\sqrt{1+k^2}}dk =-\frac12\int_{0}^{1}\frac{\text{arcsinh}{\>k}}{\text{arccsch}\>{k}}d\left(\text{arccsch}^2{k}\right) \\ =&- \frac12\text{arcsinh}\>k \>\text{arccsch}\>k\bigg|_0^1 +\frac12\int_0^1 \frac{\text{arcsinh}\>k+k\>\overset{k\to1/k}{\text{arccsch}{\>k}}}{k\sqrt{1+k^2}}dk\\ =& - \frac12\text{arcsinh}(1)\>\text{arccsch}(1) +\frac12\int_0^\infty I’(k)dk \\ = &- \frac12\text{arcsinh}^2(1)+\frac{\pi^2}8 \end{align}