Problem: Suppose that $a \in \mathbb{C}$ and that $ Re(a) \in (-1,1)$. Evaluate $\int_{-\infty}^{\infty} \frac{e^{ax}}{\cosh x} dx$ by considering the rectangular contour with vertices $\pm R$, $\pm R + \pi i$

My attempt: Let $f(z)=\cfrac{e^{az}}{\cosh z}$. Then $f$ has a simple pole at $z=\cfrac{\pi}{2}i$ and $res(f ;z=\cfrac{\pi}{2}i)=\cfrac{e^{\frac{\pi}{2}ai}}{i}$

So we have $\int_{\Gamma} f(z) dz=2\pi e^{\frac{\pi}{2}ai}$ where $\Gamma$ is the suggested rectangular contour.

Now I need to show that contributions from the sides of the rectangle vanishes as $R \rightarrow \infty$ but I keep getting wrong estimates: my upper bound for sides integral does not converge to 0 (and thats my only problem I guess).

How should I proceed?

Any helps appreciated


1 Answer 1


Let's handle the right side contribution, where $\gamma(t)=R+t\pi i$. $$ \int_{0}^{1}\frac{2e^{a(R+t\pi i)}}{e^{R+t\pi i}+e^{-R-t\pi i}}\ dt $$ By the reverse triangle inequality, $|e^{R+t\pi i}+e^{-R-t\pi i}|\geq |e^{R+t\pi i}|-|e^{-R-t\pi i}|$. Consequently we have the bound $$ \left|\frac{e^{a(R+t\pi i)}}{e^{R+t\pi i}+e^{-R-t\pi i}}\right|\leq \frac{e^{\Re(a)R-\Im(a)t\pi}}{e^R-e^{-R}}\leq \frac{e^{\Re(a)R+|\Im(a)|\pi}}{e^R-e^{-R}}. $$ Therefore we have the integral bound $$ \left|\int_{0}^{1}\frac{2e^{a(R+t\pi i)}}{e^{R+t\pi i}+e^{-R-t\pi i}}\ dt\right|\leq \frac{2e^{\Re(a)R+|\Im(a)|\pi}}{e^R-e^{-R}}. $$ Letting $R\to \infty$, the right side tends to 0 since $\Re(a)<1$. The same argument works for the left side. (Don't forget to do the top side, but it sounded like you got it covered from your question.)

  • $\begingroup$ Thanks a lot, but I'm not sure how you got the second inequality: shouldn't modulus of numerator be $exp(Re(a)R - Im(a) t \pi$ (although it doesn't make a real difference)? $\endgroup$
    – user160738
    Jan 19, 2015 at 7:27
  • $\begingroup$ Thanks for the catch, I forgot that $a$ was allowed to be complex (since I'm so used to such parameters being real). $\endgroup$
    – pre-kidney
    Jan 19, 2015 at 7:48

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