3
$\begingroup$

Just was stuck by "A remarkable Definite Integral"

From formula (3), if I substitute $a=1$, then

\begin{align} \int_{-\infty}^{\infty}\frac{2e^{-(x^2+i\pi x)t}}{e^x+e^{-x}}dx=\pi e^{-\frac{\pi^2}{4}t} \end{align}

But I couldn't proceed further. Would someone can give hints on what tricks can be used to solve this integral?

EDIT reminded by the comments that I have limited myself through converting $\cosh$ to $e^x+e^{-x}$ and then missed possibility of the pole $\frac{i\pi}{2}$.

$\endgroup$
1
  • 2
    $\begingroup$ Just consider the integral in the complex plane along the closed contour $-R\to R\to R+\pi i\to -R+\pi i\to -R;\,R\to\infty$. Side integrals tend to zero, and you get $$I=\oint\frac{e^{-t(z^2-\pi iz)}}{\cosh z}dz=\int_{-\infty}^\infty\frac{e^{-t(x^2-\pi ix)}}{\cosh x}dx+\int_{-\infty}^\infty\frac{e^{-t(x^2+\pi ix)}}{\cosh x}dx=2\int_{-\infty}^\infty\frac{e^{-t(x^2+\pi ix)}}{\cosh x}dx$$ On the other hand, having a single simpe pole inside the rectangular contour $$I=2\pi i\underset{z=\frac{\pi i}2}{\operatorname{Res}}\frac{e^{-t(z^2-\pi iz)}}{\cosh z}=2\pi e^{-\frac{\pi^2t}4}$$ $\endgroup$
    – Svyatoslav
    Commented Mar 30 at 18:35

0

Reset to default

You must log in to answer this question.

Browse other questions tagged .