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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

NoName
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Ricardo770
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3 Answers3

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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}

Quanto
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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 $$J=\sum_{n\ge0}\frac{(-1)^n2^{2n}}{(2n+1)^2\binom{2n}{n}}=\frac{\pi^2}{8}-\frac12\operatorname{arcsinh}^2(1).$$

Rócherz
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clathratus
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Slight twist on your second method that produces an equivalent form in terms of the dilogarithm :

$$\newcommand{\dilog}[1]{\operatorname{Li}_2\left(#1\right)} \\ \begin{align*} & \int_0^{\tfrac\pi2} \arctan \left(\sin x\right) \, dx \\ &= \int_0^1 \frac{\arctan y}{\sqrt{1-y^2}} \, dy \tag1 \\ &= \int_0^1 \left[\int_0^1 \frac{y}{1+y^2z^2} \, dz\right] \, \frac{dy}{\sqrt{1-y^2}} \\ &= \int_0^1 \left[\int_0^1 \frac y{\sqrt{1-y^2} \left(1+z^2y^2\right)} \, dy\right] \, dz \\ &= \int_0^1 \left[\int_1^\infty \frac w{w^4+\left(4z^2+2\right)w^2+1} \, dw\right] \, dz \tag2 \\ &= \frac12 \int_0^1 \frac{\ln\left(2z^2+1+2z\sqrt{z^2+1}\right)}{z \sqrt{z^2+1}} \, dz \\ &= \frac1{2\sqrt2} \int_0^{\ln\left(3+2\sqrt2\right)} \frac{v\coth\frac v2}{\sqrt{\cosh v+1}} \, dv \tag3 \\ &= \frac{\pi^2}4 + \ln\left(\sqrt2-1\right)\ln\left(\sqrt2+1\right) + \dilog{1-\sqrt2} - \dilog{\sqrt2-1} \\ &= \frac{\pi^2}8 + \ln\left(\sqrt2-1\right)\ln\left(\sqrt2+1\right)+\frac12\ln^2\left(\sqrt2+1\right) \tag4 \\ &= \boxed{\frac{\pi^2}8 + \frac12 \ln\left(\sqrt2+1\right) \ln\left(\sqrt2-1\right)} \end{align*}$$

  • $(1)$ Substitute $y=\sin x$
  • $(2)$ Substitute $w=\dfrac{1-\sqrt{1-y^2}}y \implies y=\dfrac{2w}{1+w^2}$
  • $(3)$ Substitute $z=\sqrt{\dfrac{\cosh v-1}2} \implies v=\ln\left(2z^2+1+2z\sqrt{z^2+1}\right)$
  • $(4)$ Evaluate using W|A's antiderivative, observing that $\sqrt{3\pm2\sqrt2}=\sqrt2\pm1$
  • $(5)$ See identity $(19)$ here
user170231
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