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I was wondering what methods people knew of to solve the following definite integral? I have found a method using Feynman's Trick (see below) but am curious as to whether there are other Feynman's Tricks and/or Methods that can be used to solve it:

$$ I = \int_{0}^{\frac{\pi}{2}} \frac{\arctan\left(\sin(x)\right)}{\sin(x)}\:dx$$

My method:

Let

$$ I(t) = \int_{0}^{\frac{\pi}{2}} \frac{\arctan\left(t\sin(x)\right)}{\sin(x)}\:dx$$

Thus,

\begin{align} I'(t) &= \int_{0}^{\frac{\pi}{2}} \frac{\sin(x)}{\left(t^2\sin^2(x) + 1\right)\sin(x)}\:dx = \int_{0}^{\frac{\pi}{2}} \frac{1}{t^2\sin^2(x) + 1}\:dx \\ &= \left[\frac{1}{\sqrt{t^2 + 1}} \arctan\left(\sqrt{t^2 + 1}\tan(x) \right)\right]_{0}^{\frac{\pi}{2}} = \sqrt{t^2 + 1}\frac{\pi}{2} \end{align}

Thus

$$I(t) = \frac{\pi}{2}\sinh^{-1}(t) + C$$

Now

$$I(0) = C = \int_{0}^{\frac{\pi}{2}} \frac{\arctan\left(0\cdot\sin(x)\right)}{\sin(x)}\:dx = 0$$

Thus

$$I(t) = \frac{\pi}{2}\sinh^{-1}(t)$$

And finally,

$$I = I(1) = \int_{0}^{\frac{\pi}{2}} \frac{\arctan\left(\sin(x)\right)}{\sin(x)}\:dx = \frac{\pi}{2}\sinh^{-1}(1) = \frac{\pi}{2}\ln\left|1 + \sqrt{2}\right|$$

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    $\begingroup$ Accepting an answer will take your question off of the "unanswered questions" page, and you probably don't want to do that until you've seen a wider variety of answers. $\endgroup$ – Frpzzd Dec 4 '18 at 0:09
  • $\begingroup$ @Frpzzd - Thanks again for your info. I have removed the "answered" status for the moment. $\endgroup$ – DavidG Dec 4 '18 at 0:17
  • $\begingroup$ I know I am a bit late in the game, but I just found an answer which I'm pretty proud of :) $\endgroup$ – clathratus Jan 12 at 23:22
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$$\begin{align} \int_0^{\pi/2}\frac{\arctan \sin(x)}{\sin(x)}dx &=\int_0^{\pi/2}\frac{1}{\sin(x)}\sum_{n=0}^\infty \frac{(-1)^n \sin^{2n+1}(x)}{2n+1}dx\\ &=\sum_{n=0}^\infty \frac{(-1)^n}{2n+1} \int_0^{\pi/2}\sin^{2n}(x)dx\\ &=\frac{\pi}{2}+\frac{\pi}{2}\sum_{n=1}^\infty \frac{(-1)^n}{2n+1}\cdot \frac{(2n-1)!!}{(2n)!!}\\ &=\frac{\pi}{2}+\frac{\pi}{2}\sum_{n=1}^\infty \frac{(-1)^n}{2^{2n-1}(2n+1)}\cdot \binom{2n-1}{n} \\ &=\frac{\pi}{2}+\frac{\pi}{2}\cdot (\sinh^{-1}(1)-1) \\ &=\frac{\pi}{2}\ln(1+\sqrt{2}) \\ \end{align}$$

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Using the following relation: $$\frac{\arctan x}{x}=\int_0^1 \frac{dy}{1+(xy)^2} \Rightarrow \color{red}{\frac{\arctan(\sin x)}{\sin x}=\int_0^1 \frac{dy}{1+(\sin^2 x )y^2}}$$ We can rewrite the original integral as: $$I = \color{blue}{\int_{0}^{\frac{\pi}{2}}} \color{red}{\frac{\arctan\left(\sin x\right)}{\sin x}}\color{blue}{dx}=\color{blue}{\int_0^\frac{\pi}{2}}\color{red}{\int_0^1 \frac{dy}{1+(\sin^2 x )y^2}}\color{blue}{dx}=\color{red}{\int_0^1} \color{blue}{\int_0^\frac{\pi}{2}}\color{purple}{\frac{1}{1+(\sin^2 x )y^2}}\color{blue}{dx}\color{red}{dy}$$ $$=\int_0^1 \left(\frac{\arctan\left(\sqrt{1+y^2}\cdot\tan(x)\right) }{\sqrt{1+y^2}} \bigg|_0^\frac{\pi}{2}\right) dy=\frac{\pi}{2}\int_0^1 \frac{dy}{\sqrt{1+y^2}}=\frac{\pi}{2}\ln\left(1+\sqrt 2\right)$$

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$\newcommand{\bbx}[1]{\,\bbox[15px,border:1px groove navy]{\displaystyle{#1}}\,} \newcommand{\braces}[1]{\left\lbrace\,{#1}\,\right\rbrace} \newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack} \newcommand{\dd}{\mathrm{d}} \newcommand{\ds}[1]{\displaystyle{#1}} \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,} \newcommand{\ic}{\mathrm{i}} \newcommand{\mc}[1]{\mathcal{#1}} \newcommand{\mrm}[1]{\mathrm{#1}} \newcommand{\pars}[1]{\left(\,{#1}\,\right)} \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,} \newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}} \newcommand{\verts}[1]{\left\vert\,{#1}\,\right\vert}$ \begin{align} I & \equiv \int_{0}^{\pi/2}{\arctan\pars{\sin\pars{x}} \over \sin\pars{x}}\,\dd x = \int_{0}^{\pi/2}\int_{1}^{\infty}{\dd t \over t^{2} + \sin^{2}\pars{x}}\,\dd x \\[5mm] & = \int_{1}^{\infty}\int_{0}^{\pi/2}{\dd x \over \sin^{2}\pars{x} + t^{2}}\,\dd t = \int_{1}^{\infty}\int_{0}^{\pi/2}{\sec^{2}\pars{x} \over \tan^{2}\pars{x} + t^{2}\sec^{2}\pars{x}}\,\dd x\,\dd t \\[5mm] & = \int_{1}^{\infty}\int_{0}^{\pi/2}{\sec^{2}\pars{x} \over \pars{1 + t^{2}}\tan^{2}\pars{x} + t^{2}}\,\dd x\,\dd t \\[5mm] & = \int_{1}^{\infty}{1 \over \root{1/t^{2} + 1}}\int_{0}^{\pi/2} {\root{1/t^{2} + 1}\sec^{2}\pars{x} \over \pars{1/t^{2} + 1}\tan^{2}\pars{x} + 1}\,\dd x\,{\dd t \over t^{2}} \\[5mm] & = \int_{1}^{\infty}{1 \over t\root{t^{2} + 1}}\int_{0}^{\infty} {\dd x \over x^{2} + 1}\,\dd x\,\dd t = {\pi \over 2}\int_{1}^{\infty}{\dd t \over t\root{t^{2} + 1}} \\[5mm] & = {\pi \over 4}\int_{1}^{\infty}{\dd t \over t\root{t + 1}} \\[5mm] & \stackrel{t\ \mapsto\ t^{2} - 1}{=}\,\,\, {\pi \over 2}\int_{\root{2}}^{\infty}{\dd t \over t^{2} - 1} = \left.{\pi \over 4}\ln\pars{t - 1 \over t + 1}\,\right\vert_{\ \root{2}}^{\ \to\ \infty} \\[5mm] & = -\,{\pi \over 4}\,\ln\pars{\root{2} - 1 \over \root{2} + 1} = {\pi \over 4}\,\ln\pars{\bracks{\root{2} + 1}^{2}} \\[5mm] & = \bbx{{\pi \over 2}\,\ln\pars{1 + \root{2}}} \approx 1.3845 \end{align}

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  • $\begingroup$ Absolutely love this method. With respect to your method of taking the integrand and converting to an integral - do you know of any texts/examples that explain the process in detail? Thanks again for the post. $\endgroup$ – DavidG Dec 5 '18 at 11:09
  • $\begingroup$ @DavidG More or less, it's similar to "Feynmann Trick". We learn many tricks "along the way". For example, in MSE. $\endgroup$ – Felix Marin Dec 5 '18 at 16:13
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$$ I = \int_{0}^{1}\frac{\arctan x}{x\sqrt{1-x^2}}\,dx =\sum_{n\geq 0}\frac{(-1)^n}{2n+1}\int_{0}^{1}\frac{x^{2n}}{\sqrt{1-x^2}}\,dx=\frac{\pi}{2}\sum_{n\geq 0}\frac{(-1)^n}{(2n+1)}\cdot\frac{\binom{2n}{n}}{4^n}$$ is a fairly simple hypergeometric series, namely $\frac{\pi}{2}\cdot\phantom{}_2 F_1\left(\tfrac{1}{2},\tfrac{1}{2};\tfrac{3}{2};-1\right)$. Since $$ \frac{1}{\sqrt{1-x}}=\sum_{n\geq 0}\frac{\binom{2n}{n}}{4^n}x^n,\qquad \arcsin(x)=\sum_{n\geq 0}\frac{\binom{2n}{n}}{(2n+1)4^n} x^{2n+1} $$ we clearly have $I=\frac{\pi}{2}\,\text{arcsin} \color{red}{\text{h}}(1) = \color{red}{\frac{\pi}{2}\log(1+\sqrt{2})}$.


By enforcing the substitution $x\mapsto\frac{1-x}{1+x}$ (involution) and exploiting the Maclaurin series of $\frac{1}{x}\left(\frac{\pi}{4}-\arctan(1-x)\right)$ I got the mildly interesting acceleration formula

$$ \frac{\pi}{2}\log(1+\sqrt{2})=\small{\sum_{k\geq 0}(-1)^k\left[\frac{2^{6k}}{(4k+1)(8k+1)\binom{8k}{4k}}+\frac{2^{6k+2}}{(4k+2)(8k+3)\binom{8k+2}{4k+1}}+\frac{2^{6k+3}}{(4k+3)(8k+5)\binom{8k+4}{4k+2}}\right]}. $$ In this case we have that a $\phantom{}_2 F_1(\ldots,-1)$ decomposes as a linear combination of three $\phantom{}_6 F_5(\ldots,-1/4)$.

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Slightly different from @Frpzzd's answer $$I=\int_0^{\pi/2}\frac{\arctan\sin x}{\sin x}\mathrm dx$$ Recall that $$\arctan x=\sum_{n\geq0}(-1)^n\frac{x^{2n+1}}{2n+1},\qquad |x|\leq1$$ And since $\forall x\in\Bbb R ,\ \ |\sin x|\leq1$, we have that $$\arctan\sin x=\sum_{n\geq0}\frac{(-1)^n}{2n+1}\sin(x)^{2n+1},\qquad \forall x\in\Bbb R$$ So we have that $$I=\sum_{n\geq0}\frac{(-1)^n}{2n+1}\int_0^{\pi/2}\sin(x)^{2n}\mathrm dx$$ I leave it as a challenge to you to prove that $$\int_0^{\pi/2}\sin(x)^a\cos(x)^b\mathrm dx=\frac{\Gamma(\frac{a+1}2)\Gamma(\frac{b+1}2)}{2\Gamma(\frac{a+b}2+1)}$$ So $$I=\sum_{n\geq0}\frac{(-1)^n}{2n+1}\frac{\Gamma(\frac{2n+1}2)\Gamma(\frac{1}2)}{2\Gamma(\frac{2n}2+1)}$$ $$I=\frac{\sqrt\pi}2\sum_{n\geq0}\frac{(-1)^n}{2n+1}\frac{\Gamma(n+\frac{1}2)}{\Gamma(n+1)}$$ Then recall that $\frac{d}{dx}\operatorname{arcsinh}x=(1+x^2)^{-1/2}$. This function has the hypergeometric representation $$\frac{d}{dx}\operatorname{arcsinh}x=\,_1\mathrm{F}_0[1/2;;-x^2]$$ $$\frac{d}{dx}\operatorname{arcsinh}x=\sum_{n\geq0}\frac{(-1)^n(1/2)_n}{n!}x^{2n}$$ Thus $$\operatorname{arcsinh}x=\sum_{n\geq0}\frac{(-1)^n(1/2)_n}{n!}\frac{x^{2n+1}}{2n+1}$$ then recalling that $(a)_n=\frac{\Gamma(a+n)}{\Gamma(a)}$, we have $$\operatorname{arcsinh}x=\sum_{n\geq0}\frac{(-1)^nx^{2n+1}}{2n+1}\frac{\Gamma(n+\frac12)}{\Gamma(\frac12)\Gamma(n+1)}$$ $$\sqrt{\pi}\,\operatorname{arcsinh}x=\sum_{n\geq0}\frac{(-1)^nx^{2n+1}}{2n+1}\frac{\Gamma(n+\frac12)}{\Gamma(n+1)}$$ And (drum roll please)... $$I=\frac{\pi}2\operatorname{arcsinh}1$$ $$I=\frac{\pi}2\log(1+\sqrt2)$$


Extra: proving the hypergeometric identity

We start by finding the Taylor Series representation for $x^\alpha$ about $x=1$. Here $\mathrm{D}^n$ represents differentiating $n$ times wrt $x$.

It is easily shown that $$\mathrm{D}^nx^\alpha=p(\alpha,n)x^{\alpha-n}$$ Where $p(\alpha,n)=\prod_{k=1}^{n}(\alpha-k+1)$ is the falling factorial. Hence $$\mathrm{D}_{x=1}^nx^\alpha=p(\alpha,n)$$ So $$x^{\alpha}=\sum_{n\geq0}\frac{p(\alpha,n)}{n!}(x-1)^n$$ $$(1+x)^{\alpha}=\sum_{n\geq0}\frac{p(\alpha,n)}{n!}x^n$$ Then using the identity $$p(\alpha,n)=(-1)^n(-\alpha)_n$$ with $(x)_n=\frac{\Gamma(x+n)}{\Gamma(x)}$, we have that $$(1+x)^\alpha=\,_1\mathrm{F}_0[-\alpha;;-x]$$ $$(1+x^2)^{-1/2}=\,_1\mathrm{F}_0[1/2;;-x^2]$$ As desired.

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    $\begingroup$ Nice solution. Especially enjoy the hypergeometric identity component. I need to spend more time focussing on these functions! I've been living in the land of Polygamma Functions lately :-) $\endgroup$ – DavidG Jan 12 at 23:41
  • $\begingroup$ Thank you :) I'm finishing my stay in polylogarithm land and working my way towards polygamma land. $\endgroup$ – clathratus Jan 12 at 23:44

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