I have an Integrand involving two exponential terms:

$$ \int_{0}^{\infty} \frac{\exp(x^2)}{(1+\exp(x^2))^2} dx $$

I would like to know what is the best way to integrate such a function without blowing it up?

What if $x^2$ is replaced by two variables $(x^2 + y^2)$ and we have a double integral?
Will the method of integration remain the same?

I use python and matlab for calculations.

Thanks

Notation fixed.

  • 2
    i think there is no formula in the known elementary functions – Dr. Sonnhard Graubner Sep 12 '16 at 15:27
  • @Dr.SonnhardGraubner what can be the best way for numerical integration? – user3840530 Sep 12 '16 at 15:35
  • 1
    @Dr.SonnhardGraubner: if $\zeta$ is considered an elementary function, you are wrong. – Jack D'Aurizio Sep 12 '16 at 15:58
  • @JackD'Aurizio, isn't $\zeta$ not an elementary function because it's an infinite sum? – tilper Sep 12 '16 at 16:35
  • @tilper: also $\sin(x)$ is an infinite sum, namely $$\sum_{n\geq 0}\frac{(-1)^n x^{2n+1}}{(2n+1)!}.$$ – Jack D'Aurizio Sep 12 '16 at 16:42
up vote 4 down vote accepted

You integral can be written as $$I=\frac{1}{4}\int_{0}^{+\infty}\frac{dx}{\cosh^2\left(\frac{x^2}{2}\right)}\,dx=\frac{1}{4\sqrt{2}}\int_{0}^{+\infty}\frac{dz}{\sqrt{z}\cosh^2(z)}=\frac{1}{8\sqrt{2}}\int_{0}^{+\infty}\frac{\tanh(z)}{z^{3/2}}\,dz $$ and due to the identity $$ \tanh(z) = \sum_{n\geq 0}\frac{8z}{(2n+1)^2 \pi^2+ 4z^2} $$ that follows from considering the logarithmic derivative of the Weierstrass product for the $\cosh$ function, we have: $$ I = \frac{1}{\sqrt{2}}\sum_{n\geq 0}\int_{0}^{+\infty}\frac{dz}{\sqrt{z}\left[(2n+1)^2 \pi^2+ 4z^2\right]}=\frac{1}{2\sqrt{2\pi}}\sum_{n\geq 0}\frac{1}{(2n+1)^{3/2}}$$ hence: $$ \boxed{I = \color{red}{\frac{2\sqrt{2}-1}{8\sqrt{\pi}}\,\zeta\left(\frac{3}{2}\right)}\approx 0.336859119428877} $$

For the two-variables integral you may ust switch to polar coordinates and apply the same technique, leading to a value of $\color{red}{\frac{\pi}{8}}$.

  • nice Jack, i tried ipb and somehow everything became divergent...^^ (+1) – tired Sep 12 '16 at 16:00
  • @tired: I tried that too, but soon after realized that the series expansion for $\tanh$ had to lead to some series solution. The only wonder is that such a series is a value of the $\eta$ function :D – Jack D'Aurizio Sep 12 '16 at 16:09
  • @JackD'Aurizio Thanks for the solution. It does give me some insight in the integral. Is cosh the best way to go about solving such an integral numerically? – user3840530 Sep 13 '16 at 14:33
  • @user3840530: you may use your favourite acceleration method to compute $$\zeta\left(\frac{3}{2}\right)=\sum_{n\geq 1}\frac{1}{n^{3/2}}\approx 2.612375 $$ – Jack D'Aurizio Sep 13 '16 at 14:34

$$ \begin{align} \int_0^\infty\frac{e^{x^2}}{\left(1+e^{x^2}\right)^2}\,\mathrm{d}x &=\int_0^\infty\frac{e^{-x^2}}{\left(1+e^{-x^2}\right)^2}\,\mathrm{d}x\\ &=\frac12\int_0^\infty\frac{e^{-x}}{(1+e^{-x})^2}\frac{\mathrm{d}x}{\sqrt{x}}\tag{1} \end{align} $$ Consider $$ \begin{align} &\int_0^\infty\frac{e^{-x}}{(1+e^{-x})^2}x^{\alpha-1}\mathrm{d}x\tag{2}\\ &=\int_0^\infty\left(e^{-x}-2e^{-2x}+3e^{-3x}-\dots\right)x^{\alpha-1}\,\mathrm{d}x\tag{3}\\ &=\Gamma(\alpha)\left(1-2^{1-\alpha}+3^{1-\alpha}-\dots\right)\tag{4}\\[6pt] &=\Gamma(\alpha)\zeta(\alpha-1)\left(1-2^{2-\alpha}\right)\tag{5} \end{align} $$ The integral in $(2)$ is analytic for $\mathrm{Re}(\alpha)\gt0$. For $\mathrm{Re}(\alpha)\gt1$ the sum in $(4)$ converges and takes the value in $(5)$. Since the functions in $(2)$ and $(5)$ are analytic and equal for $\mathrm{Re}(\alpha)\gt1$, they must be equal for $\mathrm{Re}(\alpha)\gt0$.

Therefore, with $\alpha=\frac12$, we get $$ \begin{align} \int_0^\infty\frac{e^{x^2}}{\left(1+e^{x^2}\right)^2}\,\mathrm{d}x &=\frac{\sqrt\pi}2\zeta\left(-\tfrac12\right)\left(1-\sqrt8\right)\\ &\doteq0.336859119428876991346\tag{6} \end{align} $$


A note on computing $\boldsymbol{\zeta\!\left(-\frac12\right)}$

We can't use the standard series $$ \zeta(s)=\sum_{n=1}^\infty n^{-s}\tag{7} $$ to compute $\zeta\!\left(-\tfrac12\right)$ because the series in $(7)$ only converges for $s\gt1$.

In this answer, analytic continuation is used to show that $$ \zeta\!\left(-\tfrac12\right)=\lim_{n\to\infty}\left(\sum_{k=1}^n\sqrt{k}\,-\tfrac23n^{3/2}-\tfrac12n^{1/2}\right)\tag{8} $$ The convergence of $(8)$ is very slow; the error is approximately $\frac1{24\sqrt{n}}$. However, by using $8$ terms from the Euler-Maclaurin Sum Formula, the error is reduced to $\frac{52003}{100663296}n^{-25/2}$.

Thus, using $n=1000$, we get $$ \zeta\!\left(-\tfrac12\right)=-0.2078862249773545660173067253970493022262\dots\tag{9} $$

\begin{align} \left(1 + a\mathrm{e}^{x^{2}}\right)^{-1} &= \frac{1}{a}\mathrm{e}^{-x^{2}} \frac{1}{1+\frac{1}{a\mathrm{exp}(x^{2})}} \\ &= \frac{1}{a}\mathrm{e}^{-x^{2}} \sum\limits_{n=0}^{\infty} (-1)^{n} \left( \frac{1}{a}\mathrm{e}^{-x^{2}}\right)^{n} \\ &= \sum\limits_{n=0}^{\infty} (-1)^{n} \frac{1}{a^{n+1}} \mathrm{e}^{-(n+1)x^{2}} \\ \end{align}

Now integrate the exponential function \begin{equation} \int\limits_{0}^{\infty} \mathrm{e}^{-(n+1)x^{2}} \mathrm{d}x = \frac{\sqrt{\pi}}{2} \frac{1}{\sqrt{n+1}} \end{equation}

Now we have \begin{equation} \int\limits_{0}^{\infty} \left(1 + a\mathrm{e}^{x^{2}}\right)^{-1} \mathrm{d}x = \frac{\sqrt{\pi}}{2} \sum\limits_{n=1}^{\infty} (-1)^{n-1} \frac{1}{n^{1/2}} \frac{1}{a^{n}} \end{equation}

Taking the derivative of both sides with respect to $a$:

\begin{equation} \int\limits_{0}^{\infty} \mathrm{e}^{x^{2}} \left(1 + a\mathrm{e}^{x^{2}}\right)^{-2} \mathrm{d}x = \frac{\sqrt{\pi}}{2} \sum\limits_{n=0}^{\infty} (-1)^{n-1} \frac{1}{n^{-1/2}} \frac{1}{a^{n+1}} \end{equation}

Taking $\lim_{a \to 1}$ \begin{align} \int\limits_{0}^{\infty} \mathrm{e}^{x^{2}} \left(1 + \mathrm{e}^{x^{2}}\right)^{-2} \mathrm{d}x &= \frac{\sqrt{\pi}}{2} \sum\limits_{n=1}^{\infty} (-1)^{n-1} \frac{1}{n^{-1/2}} \\ &= \frac{\sqrt{\pi}}{2} \eta(-1/2) \\ &\approx 0.336859119 \end{align}

$\eta(s)$ is the Dirichlet eta function.

Addendum: Response to robjohn's comment.

From the reference I provided for the Dirichlet eta function:

"The following relation holds: $$\eta(s) = (1 − 2^{1 − s})\zeta(s)$$

While the Dirichlet series expansion for the eta function is convergent only for any complex number $s$ with real part > 0, it is Abel summable for any complex number. This serves to define the eta function as an entire function."

  • You get the correct answer; however, the last series you give does not converge (the terms do not go to $0$). One can use analytic continuation to extend $\eta(s)$ for $s\le0$, but at least some hand-waving needs to done first. – robjohn Sep 14 '16 at 10:18

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