11
$\begingroup$

I need to prove that $\displaystyle\int_{-\infty}^{\infty} \dfrac{e^{-x}}{1+e^{-2\pi x}}\,dx=\dfrac{1}{2\sin\left(\frac{1}{2}\right)}$. This is an exercise from Basic Complex Analysis by Marsden and Hoffman. My attempt:

First, Marsden says that we need to consider the complex function $f(z)=\dfrac{e^{-z}}{1+e^{-2\pi z}}$ and the next curve

enter image description here

$\gamma_2=r+t\pi i$ with $t\in [0,1]$

$\gamma_3=(\pi i+r)-2tr$ with $t\in [0,1]$

$\gamma_4=(1-t)(\pi i-r)-tr$ with $t\in [0,1]$

Note that the only poles of the function $f(z)$ in the rectangle are when $z=\dfrac{i}{2}$, $z=\dfrac{3i}{2}$ and $z=\dfrac{5i}{2}$. From a direct calculation we obtain that $\text{Res}\left(f(z),\dfrac{i}{2}\right)=\dfrac{e^{-i/2}}{2\pi}$, $\text{Res}\left(f(z),\dfrac{3i}{2}\right)=\dfrac{e^{-3i/2}}{2\pi}$ and $\text{Res}\left(f(z),\dfrac{5i}{2}\right)=\dfrac{e^{-5i/2}}{2\pi}$.

After to a lot of calculations we obtain a bound for the integral over $\gamma_2$: $$\dfrac{e^{-r}}{\sqrt{(1-e^{-2\pi r})^2}}\geq \dfrac{|e^{-r-t\pi i}|}{|1+e^{-2\pi(r+t\pi i)}|}\geq 0$$Thus $$\displaystyle\int_{\gamma_2}^{}f(\gamma_2)\cdot d\gamma\leq \dfrac{e^{-r}}{\sqrt{(1-e^{-2\pi r})^2}}$$When $r\to\infty$ then $\displaystyle\int_{\gamma_2}^{}f(\gamma_2)\cdot d\gamma\to 0$

I think that is the same for $\gamma_4$ but I have troubles with $\gamma_3$. After a lot of calculations we obtain the next bound: $$\dfrac{e^{-r+2rt}}{\sqrt{1+2e^{-2\pi}\cos(-2\pi^2)+e^{-4\pi r}}}\geq \dfrac{|e^{-r+2rt-\pi i}|}{|1+e^{-2\pi(\pi i+r-2rt)}|}$$ but when $r\to\infty$ we obtain that $\dfrac{e^{-r+2rt}}{\sqrt{1+2e^{-2\pi}\cos(-2\pi^2)+e^{-4\pi r}}}\to\infty$ and I need, maybe, that this limits exists (maybe, zero).

Clearly I want the integrals over the curves because if $\gamma=\gamma_1\cup\gamma_2\cup\gamma_3\cup\gamma_4$ then $\displaystyle\int_{\gamma} f(\gamma)\cdot d\gamma=\displaystyle\int_{\gamma_1} f(\gamma_1) \cdot d\gamma_1+\displaystyle\int_{\gamma_2} f(\gamma_2) \cdot d\gamma_2+\displaystyle\int_{\gamma_3} f(\gamma_3) \cdot d\gamma_3+\displaystyle\int_{\gamma_4} f(\gamma_4) \cdot d\gamma_4$ and I want to use the Residue Theorem. But, is the rectangle the correct curve? Here an screenshot of the exercise:

enter image description here

$\endgroup$
  • 2
    $\begingroup$ It's Beta function (just a note, if anyone wants a way to check the result without complex methods) $\endgroup$ – Yuriy S May 28 '18 at 8:35
5
$\begingroup$

For your question, you will have to consider a rectangular contour with vertices $\pm R$ and $\pm R+i$. I'm guessing the authors meant by "same technique" as in a rectangular contour situated in the upper half of the contour plane. Calling our integrand as $f(z)$

$$f(z)\equiv\frac {e^{-z}}{1+e^{-2\pi z}}$$

and parametrizing about all four sides of the contour gives us

$$\begin{multline}\oint\limits_{\mathrm C}dz\, f(z)=\int\limits_{-R}^{R}dz\, f(x)+i\int\limits_0^1dy\, f(R+yi)-\int\limits_{-R}^{R}dz\, f(z+i)-i\int\limits_0^1dy\, f(-R+yi)\end{multline}$$

When we take the limit as $R\to\infty$, it can be shown that the integrals of the vertical sides (i.e the second and fourth integrals) vanish. This can be justified using the estimation lemma. Both integrals have a length of $L=1$ while their upper-bound $M$ can be computed by using the fact that $|e^{-yi}|\leq 1$ and $|e^{\pm R}|=e^{\pm R}$.

$$\begin{align*}M_1 & =\left|\,\frac {e^{-R}e^{-yi}}{1+e^{-2\pi R}e^{-2\pi yi}}\,\right|=\frac {e^{-R}}{1+e^{-2\pi R}}\xrightarrow{R\,\to\,\infty}0\\M_2 & =\left|\,\frac {e^{R}e^{-yi}}{1+e^{2\pi R}e^{-2\pi yi}}\,\right|=\frac {e^{R}}{1+e^{2\pi R}}\xrightarrow{R\,\to\,\infty}0\end{align*}$$

Taking their product and calling their arcs $\Gamma_{1}$ and $\Gamma_{2}$ respectively

$$\begin{align*} & \left|\,\int\limits_{\Gamma_{1}}dz\,\frac {e^{-z}}{1+e^{-2\pi z}}\,\right|\leq M_1L=0\\ & \left|\,\int\limits_{\Gamma_{2}}dz\,\frac {e^{-z}}{1+e^{-2\pi z}}\,\right|\leq M_2L=0\end{align*}$$

As in, the arc integrals vanish as $R\to\infty$. Now, what's left is

$$\oint\limits_{\mathrm C}dz\, f(z)=(1-e^{-i})\int\limits_{-\infty}^{\infty}dx\, f(x)$$

The contour integral is also equal to the sum of its residues inside the contour multiplied by $2\pi i$. Fortunately, there is only one residue inside at $z=i/2$. Therefore

$$\operatorname*{Res}_{z\,=\, i/2}\frac {e^{-z}}{1+e^{-2\pi z}}=\lim\limits_{z\,\to\, i/2}\frac {(z-i/2)e^{-z}}{1+e^{-2\pi z}}=\frac {e^{-i/2}}{2\pi}$$

Hence the contour integral is

$$\oint\limits_{\mathrm C}dz\, f(z)=ie^{-i/2}$$

Putting everything together and isolating our integral $I$, we get

$$\int\limits_{-\infty}^{\infty}dx\,\frac {e^{-x}}{1+e^{-2\pi x}}=\frac {ie^{-i/2}}{1-e^{-i}}\color{blue}{=\frac 12\csc\left(\frac 12\right)}$$

$\endgroup$
2
$\begingroup$

Somewhat briefly:

  1. Take the rectangle with the upper corners at $-r+i$ and $r+i$. Then you get one residue at $z=i/2$ and the integral over $\gamma_3$ (orientation right to left) $$ -\int_{-r}^r\frac{e^{-x-i}}{1+e^{-2\pi(x+i)}}\,dx=-e^{-i}\int_{-r}^r\frac{e^{-x}}{1+e^{-2\pi x}}\,dx. $$ It gives as $r\to\infty$ the equation for the integral $I$ to be calculated $$ I+0-e^{-i}I+0=2\pi i\cdot\frac{e^{-i/2}}{2\pi}. $$ To finish is straightforward.
  2. "After a lot of calculations" may be simplified as e.g. for $\gamma_2$ $$ \left|\frac{e^{-r-i\theta}}{1+e^{-2\pi(r+i\theta)}}\right|= \frac{e^{-r}|e^{-i\theta}|}{|1+e^{-2\pi r}e^{-i2\pi\theta}|}\le \frac{e^{-r}\cdot 1}{1-|e^{-2\pi r}e^{-i2\pi\theta}|}=\frac{e^{-r}}{1-e^{-2\pi r}}\to 0,\quad r\to\infty. $$
$\endgroup$

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.