How to evaluate $ \int_{0}^{\pi/2}\int_{0}^{x} \frac{1}{1 + \cot\left(t\right)}\,\mathrm{d}t \,\mathrm{d}x$?

Question

• I am stuck in a problem $$ \int_{0}^{\pi/2}\int_{0}^{x} \frac{1}{1 + \cot\left(t\right)}\,\mathrm{d}t \,\mathrm{d}x$$

• Using the convolution theorem I changed the double integral into a single integral \begin{align} &\int_\limits 0^{\pi/2} \left(\frac{\pi}{2}-x\right)\frac{\sin\left(x\right)} {\sin\left(x\right) + \cos\left(x\right)}\,\mathrm{d}x = \int_{0}^{\pi/2}\frac{x\cos\left(x\right)} {\sin\left(x\right) +\cos\left(x\right)}\,\mathrm{d}x \\[3mm] = &\ \int_{0}^{\pi/2}\frac{x}{1 + \tan\left(x\right)}\,\mathrm{d}x = \int_{0}^\infty\frac{\tan^{-1}\left(\theta\right)}{\left(1 + \theta\right)\left(1 + \theta^{2}\right)}\,\mathrm{d}\theta \end{align}

• [Substituted $\tan\left(x\right) = \theta$]. Using integration by part $$ v =\frac{\tan^{-1}\left(\theta\right)}{1+\theta^2} \quad\mbox{and}\quad u=\frac{1}{1+\theta}, $$ I got the following integral as $ \displaystyle\frac{1}{2}\int_{0}^\infty \left[\frac{\tan^{-1}\left(\theta\right)}{1 + \theta}\right]^{2}\,\mathrm{d}\theta $.

I am not able to evaluate that.

Answer 1

$$\begin{align} \int_0^{\pi/2}\int_0^x \frac1{1+\cot(t)}\,dt\,dx&=\int_0^{\pi/2} \frac{\pi/2-t}{1+\cot(t)}\,dt\tag1\\\\ &=\int_0^{\pi/2}\frac{(\pi/2-t) \sin(t)}{\sin(t)+\cos(t)}\,dt\tag2\\\\ &=\int_0^{\pi/2}\frac{(\pi/2-t)\sin(t)}{\sqrt{2}\cos(t-\pi/4)}\,dt\tag3\\\\ &=\int_{-\pi/4}^{\pi/4}\frac{(\pi/4-t)\sin(t+\pi/4)}{\sqrt{2}\cos(t)}\,dt\tag4\\\\ &=\frac12\int_{-\pi/4}^{\pi/4}\frac{(\pi/4-t)\left(\sin(t)+\cos(t)\right)}{\cos(t)}\,dt\tag5\\\\ &=\frac12 \int_{-\pi/4}^{\pi/4}\left(\frac\pi4 -t\tan(t)\right)\,dt\tag6\\\\ &=\frac{\pi^2}{16}-\int_0^{\pi/4}t\tan(t)\,dt\tag7\\\\ &=\frac{\pi^2}{16}-\left(\frac12 G-\frac{\pi}{8}\log(2)\right)\tag8 \end{align}$$

where $G$ is Catalan's Constant.

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NOTES:

1. Changed order of integration and carried out the resulting inner integral.
2. Used the equality $\frac1{1+\cot(t)}=\frac{\sin(t)}{\sin(t)+\cos(t)}$.
3. Used the identity $\sin(t)+\cos(t)=\sqrt{2}\cos(t-\pi/4)$.
4. Enforced the substitution $t\mapsto t+\pi/4$.
5. Expanded $\sin(t+\pi/4)=\frac{\sqrt 2}{2}(\sin(t)+\cos(t))$.
6. Exploited even and odd symmetries of integrand.
7. Carried out integral of $\pi/4$

To arrive at $(8)$, we integrate by parts with $u=t$ and $v=-\log(\cos(t))$ to find

$$\int_0^{\pi/4}t\tan(t)\,dt=\frac\pi8+\int_0^{\pi/4}\log(\cos(t))\,dt\tag9$$

Next using the Fourier series $\log(\cos(t))=-\log(2)+\sum_{k=1}^\infty \frac{(-1)^{k-1}\cos(2kt)}{k}$ in $(9)$ and integrating term by term reveals

$$\begin{align} \int_0^{\pi/4}\log(\cos(t))\,dt&=-\frac\pi4 \log(2)+\frac12\sum_{k=1}^\infty \frac{(-1)^{k-1}}{(2k-1)^2}\\\\ &=-\frac\pi4 \log(2)+\frac12 G\tag{10} \end{align}$$

Substituting $(10)$ into $(9)$, we find that

$$\int_0^{\pi/4}t\tan(t)\,dt=-\frac\pi8+\frac12 G$$

Putting it all together yields the coveted result

$$\int_0^{\pi/2}\int_0^x \frac1{1+\cot(t)}\,dt\,dx=\frac{\pi^2}{16}-\left(\frac12 G-\frac{\pi}{8}\log(2)\right)$$

Answer 2

A bit late

I want to start with OP’s last integral

$$ \displaystyle I\int_{0}^{\pi/2}\int_{0}^{x} \frac{1}{1 + \cot\left(t\right)}\,\mathrm{d}t \,\mathrm{d}x =\frac{1}{2}\int_{0}^\infty \left[\frac{\tan^{-1}\left(\theta\right)}{1 + \theta}\right]^{2}\,\mathrm{d}\theta. $$

Via integration by parts, we have $$ \begin{aligned} I & =-\frac{1}{2} \int_0^{\infty}\left(\tan ^{-1} \theta\right)^2 d\left(\frac{1}{1+\theta}\right) \\ & =- \underbrace{ \frac{1}{2}\left[\frac{\left(\tan ^{-1} \theta\right)^2}{1+\theta}\right]_0^{\infty}}_{=0} +\int_0^{\infty} \frac{\tan ^{-1} \theta}{(1+\theta)\left(1+\theta^2\right)} d \theta \end{aligned} $$ Let $\theta=\tan x$, then $$ I=\int_0^{\frac{\pi}{2}} \frac{x}{1+\tan x} d x $$ Observing that $$ \begin{aligned} \int \frac{1}{1+\tan x} d x = & \frac{1}{2} \int \frac{(\cos x+\sin x)+(\cos x-\sin x)}{\cos x+\sin x} d x \\ = & \frac{1}{2}\left[\int d x+\int \frac{d(\sin x+\cos x)}{\cos x+\sin x}\right] \\ = & \frac{1}{2}[1+\ln (\cos x+\sin x)]+C \end{aligned} $$ Via integration by parts again, we get $$ \begin{aligned} I =&\frac{1}{2} \int_0^{\frac{\pi}{2}} x d(x+\ln (\sin x+\cos x)) \\ = & \frac{1}{2}\left([x(x+\ln (\sin x+\cos x))]_0^{\frac{\pi}{2}}-\int_0^{\frac{\pi}{2}}[x+\ln (\sin x+\cos x)] d x\right) \\ = & \frac{1}{2}\left[\frac{\pi^2}{8}-\int_0^{\frac{\pi}{2}} \ln (\sin x+\cos x) d x\right]\\=& =\frac{1}{2}\left[\frac{\pi^2}{8}-\left(G-\frac{\pi}{4} \ln 2\right)\right]\cdots (*)\\=& \frac{\pi^2}{16}-\frac{G}{2}+\frac{\pi}{8} \ln 2 \end{aligned} $$ Note of (*)

$$ \begin{aligned} \int_0^{\frac{\pi}{2}} \ln (\sin x+\cos x) d x = & \frac{1}{2} \int_0^{\frac{\pi}{2}} \ln (1+\sin 2 x) d x \\ = & \frac{1}{4} \int_0^\pi \ln (1+\sin x) d x\\=&G-\frac{\pi}{4} \ln 2 \end{aligned} $$

where the last step using the result.

Answer 3

Since on $[0,\frac\pi 2]$ we have $\sin(x)+\cos(x)=\sqrt{2}\sin(x+\frac\pi 4)\ge 0$ then:

$\displaystyle \begin{align}\int_0^{\frac \pi 2}\int_0^x\frac {\sin(t)}{\sin(t)+\cos(t)}\,dt\,dx&=\frac 12\int_0^{\frac \pi 2}\bigg(x-\ln(\sin(x)+\cos(x))\bigg)\,dx\\\\&=\frac{\pi^2}{16}-\frac 12\frac\pi 2\ln(\sqrt{2})-\frac 12\int_0^{\frac \pi 2}\ln(\sin(x+\frac\pi 4))\,dx\\\\&=\frac{\pi^2}{16}-\frac{\pi\ln(2)}{8}-\int_0^{\frac \pi 4}\ln(\cos(x))\,dx\end{align}$

Last one is obtained by change $x-\frac\pi 4$ and then parity of $\cos(x)$, and its value is: $\frac K2-\frac{\pi\ln(2)}4$.

Answer 4

$\newcommand{\bbx}[1]{\,\bbox[15px,border:1px groove navy]{\displaystyle{#1}}\,} \newcommand{\braces}[1]{\left\lbrace\,{#1}\,\right\rbrace} \newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack} \newcommand{\dd}{\mathrm{d}} \newcommand{\ds}[1]{\displaystyle{#1}} \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,} \newcommand{\ic}{\mathrm{i}} \newcommand{\mc}[1]{\mathcal{#1}} \newcommand{\mrm}[1]{\mathrm{#1}} \newcommand{\on}[1]{\operatorname{#1}} \newcommand{\pars}[1]{\left(\,{#1}\,\right)} \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,} \newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}} \newcommand{\verts}[1]{\left\vert\,{#1}\,\right\vert}$ \begin{align} &\bbox[5px,#ffd]{\int_{0}^{\pi/2}\int_{0}^{x} {\dd t\,\dd x \over 1 + \cot\pars{t}}} = \int_{0}^{\pi/2}{1 \over 1 + \cot\pars{t}}\int_{t}^{\pi/2} \dd x\,\dd t \\[5mm] = &\ \int_{0}^{\pi/2}{\pi/2 - t \over 1 + \cot\pars{t}}\,\dd t \,\,\,\stackrel{t\ \mapsto\ \pi/2 - t}{=}\,\,\, \int_{0}^{\pi/2}{t \over 1 + \tan\pars{t}}\,\dd t \\[5mm] = &\ \int_{-\pi/4}^{\pi/4}{t + \pi/4 \over 1 + \bracks{\tan\pars{t} + 1}/\bracks{1 - \tan\pars{t}}}\,\dd t \\[5mm] = &\ {1 \over 2}\int_{-\pi/4}^{\pi/4}\pars{t + {\pi \over 4}} \bracks{1 -\tan\pars{t}}\,\dd t \\[5mm] = &\ -\,\ \underbrace{{1 \over 2}\int_{-\pi/4}^{\pi/4}t\tan\pars{t}\,\dd t} _{\ds{{1 \over 2}\,G - {1 \over 8}\,\pi\ln\pars{2}}}\ +\ \underbrace{{1 \over 2}\int_{-\pi/4}^{\pi/4}{\pi \over 4}\,\dd t} _{\ds{\pi^{2} \over 16}} \\[5mm] = &\ \bbx{-\,{1 \over 2}\,G + {1 \over 8}\,\pi\ln\pars{2} + {\pi^{2} \over 16}} \\ & \end{align}

User ${\tt @Mark Viola}$ already evaluated the last integral.

Answer 5

Let's evaluate the inner integral first. For this, let $$I_1=\int_0^x\frac{1}{1+\cot(t)}~\mathrm dt=\int_0^x\frac{\sin(t)}{\sin(t)+\cos(t)}~\mathrm dt;~I_2=\int_0^x\frac{\cos(t)}{\sin(t)+\cos(t)}~\mathrm dt\tag{1}$$ Clearly, $I_1+I_2=x$. And, $$I_1-I_2=-\int_0^x\frac{\cos(t)-\sin(t)}{\sin(t)+\cos(t)}~\mathrm dt=-\int_0^x\frac{1}{\sin(t)+\cos(t)}~\mathrm d\left(\sin t+\cos t\right)=-\ln\left(\sin (x)+\cos( x)\right)$$

So, $$I_1=\int_0^x\frac{1}{1+\cot(t)}~\mathrm dt=\frac12\left(x-\ln\left(\sin(x)+\cos(x)\right)\right)=\frac x2-\frac12\ln\left(\sqrt2\sin\left(x+\frac\pi4\right)\right)\tag{2}$$

The outer integral of the log sine term reminds me of the Clausen function of order $2$. So, let's evaluate it by substituting $u=2\left(x+\frac\pi4\right)$ so that $2~\mathrm dx=\mathrm du$ and $u=\frac\pi2$ to $\frac{3\pi}2$ as $x=0$ to $\frac\pi2$.

$$\begin{aligned}\int_0^{\frac\pi2}\ln\left(\sqrt2\sin\left(x+\frac\pi4\right)\right)~\mathrm dx&=\int_0^{\frac\pi2}\ln\left(\frac{1}{\sqrt2}\cdot2\sin\left(x+\frac\pi4\right)\right)\\&=-\frac12\ln2\cdot\frac\pi2+\frac12\int_{\frac\pi2}^{\frac{3\pi}2}\ln\left(2\sin\frac u2\right)~\mathrm du\\&=-\frac{\pi}{4}\ln2-\frac12\operatorname{Cl}_2\left(\frac{3\pi}2\right)+\frac12\operatorname{Cl}_2\left(\frac\pi2\right)\qquad\text{Clausen function}\\&=-\frac{\pi}4\ln2+\frac G2+\frac G2\qquad\text{relation to Catalan's constant}~ G\\&=G-\frac\pi4\ln2 \end{aligned}$$

Now, $$\begin{aligned}\int_{0}^{\frac\pi2}\int_{0}^{x} \frac{1}{1 + \cot\left(t\right)}\,\mathrm{d}t \,\mathrm{d}x&=\frac12\int_0^{\frac\pi2}x~\mathrm dx-\frac12\int_0^{\frac\pi2}\ln\left(\sqrt2\sin\left(x+\frac\pi4\right)\right)~\mathrm dx\\&=\frac12\cdot\frac12\left(\frac\pi2\right)^2+\frac\pi8\ln2-\frac G2\\&=\frac{\pi^2}{16}+\frac\pi8\ln2-\frac G2 \end{aligned}$$

Hence, $$\color{blue}{\boxed{~\int_{0}^{\frac\pi2}\int_{0}^{x} \frac{1}{1 + \cot\left(t\right)}\,\mathrm{d}t \,\mathrm{d}x=\frac{\pi^2}{16}+\frac\pi8\ln2-\frac G2=\frac1{16}\left(\pi^2+2\pi\ln2-8G\right)~}}$$

where $G=0.915965594\ldots$ is Catalan's constant.

Answer 6

To evaluate $\int_0^{\pi/2} \int_0^{x} \frac 1 {1+\cot(t)}\,dt \,dx$ first work with only the inner integral in $t$, $\int_0^x \frac 1 {1+\cot(t)}dt=\int_0^x \frac {\tan(t)}{1+\tan(t)}dt$. Change integration variable to $y$ where $y=\tan(t)$ and using $\frac {dt}{d\,\tan(t)}=\frac 1 {(\frac{d\,tan(t)}{dt})}=\frac 1 {1+\tan(t)^2}= \frac 1{1+y^2}$ the inner integral equals $\int_{\tan(0)}^{\tan(x)} \frac y {(y+1)(y^2+1)}dy$. Expanding into partial fractions the general integral is found by solving for $A,B,C$ in $\frac y{(1+y)(1+y^2)}= A/(y+1)+B/(y+i)+C/(y-i)$ or $y=A(y^2+1)+B(y-i)(y+1)+C(y+i)(y+1)$ where $i$ is the positive imaginary $\sqrt{-1}$. We find $\frac y {(1+y)(1+y^2)}=\frac {-1}{2(y+1)}+\frac 1{2(1-i)(y+i)} +\frac 1{2(1+i)(y-i)}$. Recombining the complex parts obtain $\frac {-1}{2(y+1)} +\frac{y+1}{2(y^2+1)}$. So to find the inner integral over the limits, integrate this with respect to $y$ and evaluate at $y=\tan(x)$ and subtract that at $y=\tan(0)=0$ (which is $0$) and obtain $\int_{\tan(0)}^{\tan(x)}\!\! \left ( \frac {-1}{2(y\!+\!1)}\!+\!\frac{y\!+\!1}{2(y^2\!+\!1)}\right)\!dy\!=\!-\frac {\ln(\tan(x)\!+\!1)} {2}+ \frac {\ln(\tan(x)^2\!+\!1)} {4}\!+\!\frac{\arctan(\tan(x))} 2$. So for the whole double integral we put this inner integral into the outer integral which is integrate over $x$ from $0$ to $\pi/2$. With help of maxima , which is known to be rather 'flaky' at times, the integral is rather difficult but most of the infinite parts cancel and can also save some effort by taking the real part since we know the integral must be real. The result of the general integral is $ \frac{imagpart(Li_2(i \exp(2ix)))+x\ln(2)+x^2 } 4$
where $Li_2(z)$ is the so called Dilogarithm , a special case of the polylogarithm and defined by $\int_z^{0} \frac {\ln(1-t)} {t} dt = \sum_{k=1}^{\infty} \frac {z^k}{k^2} $. Evaluation of the integral at $x=\frac \pi 2$ and subtracting that at $x=0$ gives $-\frac{ imagpart(Li_2(i))} 2+\frac {\pi \ln(2)} 8 +\frac {\pi^2}{16} $ where have used $imagpart(Li_2(-it))=-imagpart(Li_2(it))$ where $t$ is real. For evaluation of $-imagpart(Li_2(i))$ used $-\sum_{k=0}^\infty (-1)^k/(2k+1)^2$. For practical use of course had to replace the upper limit at $\infty$ by a sufficiently large enough integer according to precision desired. Also a numerical integral estimate was done by double romberg that is romberg(romberg(....) integration and agreed with the above to within 8 decimal places or more , the numerical result estimate being $ 0.431065739 $.