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I was working on an integral which I found on Quora. I simplified it a lot and ended up with this intgeral

$$\int_0^{\infty}\dfrac{\operatorname{sech}(\pi x)}{1+4x^2}\, \mathrm dx $$

I tried converting this into exponential form and using geometric series which ended up in this.

$$2\sum_{k=0}^\infty (-1)^k\int_0^\infty \dfrac{e^{-(2k+1)\pi x}}{4x^2+1}\, \mathrm dx $$

I didn't try to solve this using exponential integral, as I am not that much familiar with it.

Using Wolfram|Alpha, I figured out that this integral is equal to $\frac{1}{2}\ln{2}$. The simple answer makes me suspect if the integral is just a tricky one.

How can I evaluate this integral, using this method or any other method, except contour integration?

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  • $\begingroup$ Maybe try writing $\frac{1}{1+4x^2} $ as a geonetric series. That way you get integrals like $\int x^n e^{-x} dx $ which are related to the gamma function. $\endgroup$
    – Tavish
    Jun 20, 2021 at 15:29
  • $\begingroup$ @Tavish the integration bounds are from $0$ to $\infty $ so won't work directly , however you could just split the integral $\sum_{k=0}^\infty \int_{k}^{k+1} f(x)\mathrm{d}x$ and then use a substituiton to $x-k=u$ to get the bounds to $0$ to $1$, then maybe $\endgroup$ Jun 20, 2021 at 15:32
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    $\begingroup$ @SarthakSahoo Splitting the integral as $\int_0^{1/2}+ \int_{1/2}^{\infty} $ would work. $\endgroup$
    – Tavish
    Jun 20, 2021 at 15:50
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    $\begingroup$ A shame they are closing the question for “lack of context”. Look at all these creative answers. Fourier, residues, expansions, clever changes of variables... this is the type of post that we all benefit from. This site’s rules are way too rigid sometimes. $\endgroup$ Jun 26, 2021 at 5:27
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    $\begingroup$ @Stefan thanks for understanding. Don't know why would someone report it. The question doesn't even require more context. More context will be useless. $\endgroup$ Jun 26, 2021 at 11:42

5 Answers 5

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Integrate as follows

\begin{align} &\int_0^{\infty}\frac{\operatorname{sech}(\pi x)}{1+4x^2}\>dx\\ \overset{t=2\pi x }=& \frac\pi2 \int_{-\infty}^{\infty}\frac{e^{\frac t2}}{(e^t+1)(\pi^2+t^2)}dt\\ =& \frac\pi2 \int_{-\infty}^{\infty}\frac{e^{\frac t2}}{e^t+1}Re\left(\frac1\pi \int_0^\infty e^{-(\pi-i t)y }dy\right)dt\\ =& \frac12 Re\int_{0}^{\infty}e^{-\pi y} \left(\int_{-\infty}^\infty \frac{e^{a t}}{e^t+1}dt \right)dy \>\>\>\>\>\>\>a= iy+\frac12\\ \overset{x=e^t}=& \frac12 Re\int_{0}^{\infty}e^{-\pi y} \left(\int_{0}^\infty \frac{x^{a-1}}{x+1}dx \right)dy \\ =& \frac12Re \int_{0}^{\infty}e^{-\pi y}\pi\csc(\pi a)\,dy \\ =& \int_{0}^{\infty} \frac\pi{e^{2\pi y}+1}dy \overset{t=e^{-2\pi y}}=\frac12\int_0^1 \frac1{1+t}dt\\ =&\frac12\ln2 \end{align}

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    $\begingroup$ Straightforward. +1 $\endgroup$
    – K.defaoite
    Jun 20, 2021 at 18:13
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    $\begingroup$ Excellent approach. You will surely be featured in my answer on Quora. The way you manipulated $ \frac{1}{\pi^2+x^2}$ was really nice. $\endgroup$ Jun 20, 2021 at 18:23
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All proposed different ways of solution to the problem are really nice. In my opinion, contour integration "catches" the symmetry of this problem deeper and allows to get the closed answer to more general problem.

Let's denote $$I(a)=\int_{-\infty}^\infty\frac{dx}{(e^{\pi x}+e^{-\pi x})(a^2+x^2)}$$ Then $$I_0=\int_0^{\infty}\dfrac{\operatorname{sech}(\pi x)}{1+4x^2}\, \mathrm dx=\frac{1}{4}\int_{-\infty}^\infty\frac{dx}{(e^{\pi x}+e^{-\pi x})(\bigl(\frac{1}{2}\bigr)^2+x^2)}=\frac{1}{4}I\Bigl(\frac{1}{2}\Bigr)$$ Let's also consider $$J(a)=\int_{-\infty}^\infty\frac{\log\bigl(a^2+x^2\bigr)}{e^{\pi x}+e^{-\pi x}}dx$$

Then $I(a)=\frac{1}{2a}\frac{\partial}{\partial a}J(a)$ and

$$J(a)=2\,\Re\int_{-\infty}^\infty\frac{\log\bigl(a-ix\bigr)}{e^{\pi x}+e^{-\pi x}}dx=\frac{2}{\pi}\Re\int_{-\infty}^\infty\frac{\log\bigl(a-\frac{it}{\pi}\bigr)}{e^t+e^{-t}}dt$$ $$=\frac{2}{\pi}\Re\int_{-\infty}^\infty\frac{\log\bigl(\frac{a}{2}-\frac{it}{2\pi}\bigr)+\log2}{e^t+e^{-t}}dt=J_1(a)+J_2$$ where $$J_2=\frac{2\log2}{\pi}\int_{-\infty}^\infty\frac{dt}{e^t+e^{-t}}=\frac{2\log2}{\pi}\int_{-\infty}^\infty\frac{e^tdt}{e^{2t}+1}=\frac{2\log2}{\pi}\int_{0}^\infty\frac{dx}{x^2+1}=\log2$$ To evaluate $J_1(a)$ we follow the approach developed by Iaroslav V. Blagouchine (here)

Noting that $\log z=\log\Gamma (z+1)-\log\Gamma(z)$ $$J_1(a)=\frac{2}{\pi}\Re\int_{-\infty}^\infty\frac{\log\bigl(\frac{a}{2}-\frac{it}{2\pi}\bigr)}{e^t+e^{-t}}dt=\frac{2}{\pi}\Re\int_{-\infty}^\infty\frac{\log\Gamma\bigl(\frac{a}{2}-\frac{it}{2\pi}+1\bigr)-\log\Gamma\bigl(\frac{a}{2}-\frac{it}{2\pi}\bigr)}{e^t+e^{-t}}dt$$

Next, we choose a rectangular contour C in the complex plane enter image description here

Noting that integral along the path $(1)$ and $(2)$ wanish as $R\to\infty$ and that $e^{2\pi i+t}=e^t$, we can write: $$J_1(a)=-\frac{2}{\pi}\Re\oint_C\frac{\log\Gamma\bigl(\frac{a}{2}-\frac{iz}{2\pi}\bigr)}{e^z+e^{-z}}dz=\Re\biggl(-2\pi i \frac{2}{\pi}\operatorname*{Res}_{\binom{z=\pi i/2}{z=3\pi i/2}}\frac{\log\Gamma\bigl(\frac{a}{2}-\frac{iz}{2\pi}\bigr)}{e^z+e^{-z}}\biggr)=2\Bigl(\log\Gamma\bigl(\frac{a}{2}+\frac{3}{4}\bigr)-\log\Gamma\bigl(\frac{a}{2}+\frac{1}{4}\bigr)\Bigr)$$ $$J(a)=\int_{-\infty}^\infty\frac{\log\bigl(a^2+x^2\bigr)}{e^{\pi x}+e^{-\pi x}}dx=J_1(a)+J_2=2\Bigl(\log\Gamma\bigl(\frac{a}{2}+\frac{3}{4}\bigr)-\log\Gamma\bigl(\frac{a}{2}+\frac{1}{4}\bigr)\Bigr)+\log2$$ Taking derivative with respect to $a$ we can easily evaluate expressions like $\int_{-\infty}^\infty\frac{dx}{(e^{\pi x}+e^{-\pi x})(a^2+x^2)^n}$.

For example, $$I_0=\int_0^{\infty}\dfrac{\operatorname{sech}(\pi x)}{1+4x^2}\, \mathrm dx=\frac{1}{4}I\bigl(a=\frac{1}{2}\bigr)=\frac{1}{4}\frac{1}{2a}\frac{\partial}{\partial a}J(a)|_{a=1/2}=\frac{1}{8a}\Bigl(\Psi\bigl(\frac{a}{2}+\frac{3}{4}\bigr)-\Psi\bigl(\frac{a}{2}+\frac{1}{4}\bigr)\Bigr)|_{a=1/2}$$ $$I_0=\frac{1}{4}\Bigl(\Psi(1)-\Psi(\frac{1}{2})\Bigr)=\frac{\log2}{2}$$ where $\Psi(x)=\frac{\partial}{\partial x}\log\Gamma(x)$ - digamma function.

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    $\begingroup$ Nice solution and good reference. $\endgroup$
    – NoName
    Jun 23, 2021 at 15:48
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    $\begingroup$ You can also integrate the digamma function directly and use the recurrence relation of the digamma function. I intended to leave a comment a while ago, but I guess I forgot. $\endgroup$ Dec 4, 2021 at 23:47
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The question asks for anything "except contour integration", but actually the latter is the very first thing I would apply here. So this answer follows the contour integration approach for reference.

The integral is $\frac12\int_{-\infty}^\infty$, so that if $I_{R,N}=\int_{C_{R,N}}\frac{\operatorname{sech}\pi z}{1+4z^2}\,dz$ where, for $R>0$ and $N$ a positive integer, we take $C_{R,N}$ to be the boundary of $[-R,R]+i[0,N]$, then $\lim\limits_{R\to\infty}\lim\limits_{N\to\infty}I_{R,N}$ is twice the given integral, and it equals $2\pi i$ times the sum of the residues of the integrand at $z=i(n+1/2)$ over $n\in\mathbb{Z}_{\geqslant 0}$ ($n=0$ is a double pole, $n>0$ are simple poles): $$\operatorname*{Res}_{z=i/2}\frac{\operatorname{sech}\pi z}{1+4z^2}=\frac1{4\pi i},\quad\operatorname*{Res}_{z=i(n+1/2)}\frac{\operatorname{sech}\pi z}{1+4z^2}=\frac1{4\pi i}\frac{(-1)^{n-1}}{n(n+1)}.\quad(n>0)$$ Hence the value of the given integral is, as conjectured, $$\frac14\left[1+\sum_{n=1}^\infty(-1)^{n-1}\left(\frac1n-\frac1{n+1}\right)\right]=\frac14\big(1+\ln2-(1-\ln2)\big)=\frac{\ln2}{2}.$$

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  • $\begingroup$ Well, I had not requested for contour integration approach, because I haven't learned it yet. I knew that it would be much straightforward. $\endgroup$ Jun 20, 2021 at 18:26
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    $\begingroup$ Both solutions are really nice. I also think that contour integration provides a shortcut to more general solution (for an arbitrary parameter, for instance, or more general type of integral). I tried to use the method developed by I. Blagouchine $\endgroup$
    – Svyatoslav
    Jun 21, 2021 at 22:53
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A short solution by using the series of $\operatorname{sech}(x)$.

Since $$\operatorname{sech}(x)= 4\sum_{k=0}^{\infty}\frac{(-1)^k(2k+1)\pi}{(2k+1)^2\pi^2+4x^2}$$ and then we have $$\int_0^{\infty}\frac{\operatorname{sech}(\pi x)}{1+4x^2}dx=\frac{4}{\pi}\sum_{k=0}^{\infty}{(-1)^k(2k+1)}\int_0^{\infty}\frac{dx}{(m^2+4x^2)(1+4x^2)}=$$ where $m=2k+1$ and by partial fraction decomposition we easily can deduce the integral $$\int_0^{\infty}\frac{dx}{(m^2+4x^2)(1+4x^2)}=\frac{\pi}{4m(m+1)}=\frac{\pi}{4(2k+1)(2k+2)}$$ and hence we have $$\sum_{k=0}^{\infty}\frac{(-1)^k}{2(k+1)}=\frac{1}{2}\sum_{k=1}^{\infty}\frac{(-1)^{k-1}}{k}=\frac{\log(2)}{2}$$

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  • $\begingroup$ Where does that series representation come from? $\endgroup$
    – FShrike
    Jan 29 at 9:17
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Note that your integrand is even in $x$, so that you are really concerned with

$$ \frac{1}{2}\int_{\mathbb{R}} \operatorname{sech}(\pi x) \frac{1}{1+4x^2}\,dx.$$

If you're willing to accept Fourier transform techniques, we have

$$ \frac{1}{\sqrt{2\pi}}\int_{\mathbb{R}} e^{-ixy}\operatorname{sech}(\pi x)\,dx = \frac{1}{\sqrt{2\pi}}\operatorname{sech}\bigg(\frac{y}{2}\bigg) $$

and also

$$ \frac{1}{\sqrt{2\pi}} \int_{\mathbb{R}} e^{-ixy} \frac{1}{1+4x^2}\,dx = \frac{1}{2}\sqrt{\frac{\pi}{2}} e^{-|y|/2}. $$

Then from Parseval's theorem,

\begin{align} \frac{1}{2}\int_{\mathbb{R}} \operatorname{sech}(\pi x) \frac{1}{1+4x^2}\,dx &= \frac{1}{2} \int_{\mathbb{R}} \frac{1}{\sqrt{2\pi}} \operatorname{sech}\bigg(\frac{y}{2}\bigg) \frac{1}{2}\sqrt{\frac{\pi}{2}} e^{-|y|/2}\,dy \\ &= \frac{1}{4} \int_0^{\infty} \operatorname{sech}\bigg(\frac{y}{2}\bigg) e^{-y/2}\,dy. \end{align}

As for this integral, note that $\operatorname{sech}(x) = \frac{2}{e^x + e^{-x}}$ so that we are left to evaluate

$$ \frac{1}{2}\int_0^{\infty} \frac{e^{-y/2}}{e^{y/2} + e^{-y/2}}\,dy. $$

Let $z = e^{-y/2}$, then $dz = -\frac{1}{2} e^{-y/2}\,dy$, to give

$$ \int_0^1 \frac{1}{z^{-1}+z}\,dz = \int_0^1 \frac{z}{1+z^2}\,dz = \frac{\log(2)}{2}.$$

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