8
$\begingroup$

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

$\endgroup$
4
3
$\begingroup$

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}

$\endgroup$
0
$\begingroup$

Since $|\sin x|\le 1$, we may write $$\arctan(\sin x)=\sum_{n\ge0}\frac{(-1)^n}{2n+1}\sin(x)^{2n+1},$$ and thus $$J=\int_0^{\pi/2}\arctan(\sin x)dx=\sum_{n\ge0}\frac{(-1)^n}{2n+1}\int_0^{\pi/2}\sin(x)^{2n+1}dx.$$ The remaining integral is an easy use of the Beta function, and we have $$\int_0^{\pi/2}\sin(x)^{2n+1}dx=\frac{2^{2n}n!^2}{(2n+1)!},$$ so that $$J=\sum_{n\ge0}\frac{(-1)^n2^{2n}}{(2n+1)^2\binom{2n}{n}}.$$ Then this gives $$J=\sum_{n\ge0}\frac{(-1)^n2^{2n}}{(2n+1)^2\binom{2n}{n}}=\frac{\pi^2}{8}-\frac12\text{arcsinh}^2(1).$$

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.