There are at least two ways to show that $$\int_{-\infty}^{\infty} \frac{\cos (ax^{2}) \cosh(ax)}{\cosh( \pi x)} \ dx = \cos \left( \frac{a}{4}\right) \ , \ |a| \le \pi $$ using contour integration.

One way is to integrate $ \displaystyle f(z) = \frac{e^{iaz^{2}}e^{az}}{\cosh (\pi z)}$ around a rectangle with vertices at $z=R, z= R+i$, $z=-R+i$ and $z=-R$.

A second less obvious way is to integrate $\displaystyle g(z) = \frac{e^{iaz^{2}}}{\sinh (\pi z)}$ around a rectangle with vertices at $z= \pm R \pm \frac{i}{2}$.

But what if we replace $\cosh(ax)$ with $\cosh (bx)$?

Can $$\int_{-\infty}^{\infty} \frac{\cos(ax^{2}) \cosh(bx)}{\cosh(\pi x)} \ dx \ , \ |b| \le \pi $$ be evaluated in closed form?

Simply letting $ \displaystyle f(z) = \frac{e^{iaz^{2}} e^{bz}}{\cosh (\pi z)}$ and integrating around the first contour won't work.

And I'm interested in any approach, not necessarily one that involves contour integration.

  • $\begingroup$ I felt there should be some formula for that. I know $\int \dfrac{\cosh(bx)}{\cosh(\pi x)} = \sec(b/2)$ $\endgroup$
    – Yimin
    Commented Feb 20, 2013 at 22:43

1 Answer 1


A general approach to tackle both problems is to exploit the fact that: $$ f(x) = \frac{1}{\cosh(x\sqrt{\pi/2})}\tag{1}$$ is a fixed point for the Fourier transform $\mathcal{F}:f\to\widehat{f}$: $$ \widehat{f}(s) = \frac{1}{\sqrt{2\pi}}\int_{-\infty}^{+\infty}f(x)\,e^{-isx}\,dx.$$ If $g(x)=\cos(ax^2),h(x)=\cosh(bx)$ we have: $$ \widehat{g}(s) = \frac{1}{2\sqrt{a}}\left(\cos\frac{s^2}{4a}+\sin\frac{s^2}{4a}\right),$$ $$\widehat{h}(s) = \sqrt{\frac{\pi}{2}}\left(\delta(s-ib)+\delta(s+ib)\right)\tag{2}$$ so we have: $$ \int_{-\infty}^{+\infty}\frac{\cosh(bx)}{\cosh(\pi x)}\,dx = \frac{1}{\cos\frac{b}{2}}=\sum_{n\geq 0}\frac{(-1)^n E_{2n}}{4^n(2n)!}b^{2n}\tag{3}$$ for any $b\in\mathbb{C}$ such that $|b|<\pi$.

For instance, by differentiating $(3)$ with respect to $b$ twice, we get: $$ \int_{-\infty}^{+\infty}x^2\,\frac{\cosh(bx)}{\cosh(\pi x)}\,dx = \frac{d^2}{db^2}\frac{1}{\cos\frac{b}{2}}=\frac{3-\cos b}{8\cos^3\frac{b}{2}}\tag{4}$$ so: $$\begin{eqnarray*} \int_{-\infty}^{+\infty}\cos(ax^2)\frac{\cosh(bx)}{\cosh(\pi x)}\,dx &=& \sum_{m=0}^{+\infty}\frac{(-1)^m a^{2m}}{(2m)!}\int_{-\infty}^{+\infty}x^{4m}\frac{\cosh(bx)}{\cosh(\pi x)}\,dx\\&=& \sum_{m=0}^{+\infty}\frac{(-1)^m a^{2m}}{(2m)!}\cdot\frac{d^{4m}}{db^{4m}}\frac{1}{\cos\frac{b}{2}}\tag{5}\end{eqnarray*}$$ and now it is sufficient to exploit the identity: $$ \sec\frac{b}{2}=2\sum_{n\geq 0}(-1)^n\left(\frac{1}{b+(2n+1)\pi}-\frac{1}{b-(2n+1)\pi}\right)\tag{6}$$ that follows from the residue theorem.


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