3
$\begingroup$

How can I calculate the integral $$\int_0^{\pi /2}\frac x {\tan x} \, dx$$ with complex integration? (Contours, residue theorem, etc.) I was thinking on using the fact that $\displaystyle \tan x=\frac{e^{ix}-e^{-ix}}{i(e^{ix}+e^{-ix})}$, which implies $e^{ix}=z$. I still have not been succesful.

$\endgroup$
4
  • 1
    $\begingroup$ You can do it with real methods. Let $f(a)=\int_{0}^{0.5\pi} \frac{\arctan(a \tan(x))}{\tan x} dx$. Then $f'(a)=..$ $\endgroup$ – Ahmed S. Attaalla Aug 21 '17 at 22:04
  • 3
    $\begingroup$ HI! Yes, I know. But now I am studying complex integration and I'm only interested on this method. $\endgroup$ – otreblig Aug 21 '17 at 22:10
  • $\begingroup$ I used the substitution x = arctan u then I represented the original integral as a double integral. Don't have time now, but thanks for your attention. $\endgroup$ – otreblig Aug 21 '17 at 23:38
  • $\begingroup$ Are you sure how your statement ? Ahmed's method seems the best. $\endgroup$ – reuns Aug 22 '17 at 2:47
5
$\begingroup$

A complex-analytic solution. Here is a combination of calculus and a basic complex analysis: Perform integration by parts to sanitize the integrand:

$$ \int_{0}^{\frac{\pi}{2}} \frac{x}{\tan x} \, dx = \underbrace{ \left[ \vphantom{\int} x \log\sin x \right]_{0}^{\frac{\pi}{2}} }_{=0} - \int_{0}^{\frac{\pi}{2}} \log \sin x \, dx. $$

Also notice that, if $x \in (0,\frac{\pi}{2}]$ then $\log|\sin x| = -\log 2 + \operatorname{Re}\log(1 - e^{2ix})$. So

\begin{align*} \int_{0}^{\frac{\pi}{2}} \frac{x}{\tan x} \, dx &= - \frac{1}{2} \int_{0}^{\pi} \log |\sin x| \, dx \\ &= \frac{\pi}{2}\log 2 - \frac{1}{2} \operatorname{Re} \left( \int_{0}^{\pi} \log(1 - e^{2ix}) \, dx \right) \\ (z = e^{2ix}) \quad&= \frac{\pi}{2}\log 2 - \frac{1}{4} \operatorname{Re} \left( \oint_{|z|=1} \frac{\log(1 - z)}{iz} \, dz \right), \end{align*}

Since $z \mapsto \frac{\log(1 - z)}{iz}$ is holomorphic in the unit disc $\mathbb{D}$ and has only logarithmic singularity on the boundary $\partial \mathbb{D}$, we can still apply the Cauchy integral theorem to conclude that the integral vanishes. Therefore we have

$$ \int_{0}^{\frac{\pi}{2}} \frac{x}{\tan x} \, dx = \frac{\pi}{2}\log 2. $$


Alternative complex-analytic solution. Use the substitution $x = \arctan u$ to write

$$ \int_{0}^{\frac{\pi}{2}} \frac{x}{\tan x} \, dx = \frac{1}{2}\int_{-\infty}^{\infty} \frac{\arctan u}{u(u^2+1)} \, du. $$

This hints us that we may use some contour integration technique, but we need to resolve the branch cut of $\arctan$. Here we give one such trick. Notice that

$$ \frac{\arctan u}{u} = \int_{0}^{1} \frac{dt}{1+u^2 t^2}. $$

Plugging this back and interchanging the order of integration, we get

$$ \int_{0}^{\frac{\pi}{2}} \frac{x}{\tan x} \, dx = \frac{1}{2}\int_{0}^{1} \left( \int_{-\infty}^{\infty} \frac{du}{(u^2 t^2+1)(u^2+1)} \right) \, dt. $$

Now we can perform contour integration to obtain that

\begin{align*} \int_{-\infty}^{\infty} \frac{du}{(u^2 t^2+1)(u^2+1)} &= 2\pi i \left( \underset{u = i}{\mathrm{Res}} \, \frac{1}{(u^2 t^2+1)(u^2+1)} + \underset{u = i/t}{\mathrm{Res}} \, \frac{1}{(u^2 t^2+1)(u^2+1)} \right) \\ &= \frac{\pi}{t+1}. \end{align*}

So we have

$$ \int_{0}^{\frac{\pi}{2}} \frac{x}{\tan x} \, dx = \frac{1}{2}\int_{0}^{1} \frac{\pi}{t+1} \, dt = \frac{\pi}{2} \log 2. $$

$\endgroup$
2
  • $\begingroup$ This might be more in line with what OP was looking for, but I'd be remiss if I didn't mention that that latter integral is solvable with real methods. math.stackexchange.com/questions/2381701/… $\endgroup$ – David Bowman Aug 22 '17 at 3:47
  • 3
    $\begingroup$ @DavidBowman, I agree :) Solution using symmetry is certainly a thing that we would not want to miss. It is also interesting to see that this integral can be solved by Fourier series and harmonic analysis. $\endgroup$ – Sangchul Lee Aug 22 '17 at 3:56
2
$\begingroup$

I would like to add an alternative approach. By considering the logarithmic derivative of the Weierstrass product $$ \sin(\theta) = \theta\prod_{n\geq 1}\left(1-\frac{\theta^2}{\pi^2 n^2 }\right) \tag{1}$$ we get: $$ \theta\cot\theta = 1-\sum_{n\geq 1}\frac{2\theta^2}{\pi^2 n^2 -\theta^2} \tag{2}$$ from which: $$ \int_{0}^{\pi/2}\theta\cot\theta\,d\theta = \frac{\pi}{2}+\pi\sum_{n\geq 1}\left(1-2n\,\text{arctanh}\frac{1}{2n}\right) \tag{3}$$ where the RHS of $(3)$ can be computed through summation by parts and Stirling's approximation. Indeed, $$\begin{eqnarray*}\sum_{n=1}^{N}\left(1-n \log\frac{2n+1}{2n-1}\right) &=&N-N\log(2N+1)+\sum_{n=1}^{N-1}\log(2n+1)\\&=&N-N\log(2N+1)+\log\frac{(2N)!}{2^N N!}\\&=&\frac{\log 2-1}{2}+O\left(\frac{1}{N}\right)\tag{4}\end{eqnarray*}$$ leads to $\int_{0}^{\pi/2}\frac{x}{\tan x}\,dx = \color{blue}{\frac{\pi}{2}\log 2}$. Still another approach is to exploit Riemann sums (!!!). We have

$$ \int_{0}^{\pi/2}\theta\cot\theta\,d\theta\stackrel{IBP}{=}-\int_{0}^{\pi/2}\log\sin\theta\,d\theta=-\frac{1}{2}\int_{0}^{\pi}\log\sin\theta\,d\theta\tag{5}$$ and $$ \prod_{k=1}^{N-1}\sin\left(\frac{\pi k}{N}\right)=\frac{2N}{2^N},\tag{6} $$ so that: $$ \int_{0}^{\pi/2}\theta\cot\theta\,d\theta = -\frac{\pi}{2}\lim_{N\to +\infty}\frac{1}{N}\log\frac{2N}{2^N}=\color{blue}{\frac{\pi}{2}\log 2}.\tag{7}$$ At last, an efficient overkill: since $\Gamma(z)\,\Gamma(1-z)=\frac{\pi}{\sin(\pi z)}$, the equality between the RHS of $(5)$ and $\frac{\pi}{2}\log 2$ is a straightforward consequence of Raabe's Theorem on $\int\log\Gamma$.

$\endgroup$
1
  • $\begingroup$ This one is simple yet brilliant. +1 $\endgroup$ – Paramanand Singh Aug 22 '17 at 17:23
1
$\begingroup$

Let $z=\tan{x}$, then the integral you're looking for is $\int_{0}^{\infty}\frac{\arctan{z}}{z}\frac{1}{1+z^{2}}dz$. The integrand is an even function of $z$. This integral is then equal to $\frac{1}{2}\int_{-\infty}^{\infty}\frac{\arctan{z}}{z}\frac{1}{1+z^{2}}dz$. Use the residue theorem to evaluate this one

$\endgroup$
1
  • $\begingroup$ How do you use the residue theorem ? $\text{arctan}(z)$ has some branch points $\endgroup$ – reuns Aug 22 '17 at 2:42

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.