Hi there!

I am trying to confirm the following integral using the residue theorem but I'm stuck in the end and I don't know what I did wrong:

$\displaystyle \int_{-\infty}^{+\infty}\frac{1}{\cosh x}dx=\pi$

Here's my solution:

We chose \gamma to be the positively oriented semicircle around the origin of radius R, so we get:

$\displaystyle \gamma=\gamma_1+\gamma_2$

$\displaystyle \gamma_1=Re^{it}, t\in[0, \pi]$

$\displaystyle \gamma_2=t, t\in[-R,R]$

and we are going to apply the residue thm on the following holomorphic function:

$\displaystyle \oint_{\gamma}\frac{1}{\cosh z}dz=2\pi i\sum_k Res_{z=c_k}f(z)$

where $\displaystyle \frac{1}{\cosh z}=f(z)$ and c_k are the zeroes of coshz, i.e. $\displaystyle c_k=\frac{2k+1}{2}\pi i$

First, we show that the integral over the semicircle vanishes for big R:

Using the estimation lemma, where ||-|| denotes the supremum norm, we get

$\displaystyle ||\oint_{\gamma}\frac{1}{\cosh z}dz||_{\gamma}=2\pi R||\frac{1}{e^z+e^{-z}}||_{\gamma}$

Observe that $\displaystyle ||e^z+e^{-z}||\geq ||e^z||-||e^{-z}||=e^{Re^{it}}-e^{-Re^{it}}\geq e^R-e^{-R}$

so it follows that

$\displaystyle ||\oint_{\gamma}\frac{1}{\cosh z}dz||_{\gamma}=2\pi R||\frac{1}{e^z+e^{-z}}||_{\gamma}\leq 2\pi R\frac{1}{e^R-e^{-R}}\longrightarrow^{R\rightarrow\infty}0$ using l'Hospitals rule.

so we got: $\displaystyle \int_{\gamma_1}\frac{1}{\cosh z}dz=0$ for R very big, and

$\displaystyle \int_{\gamma_2}\frac{1}{\cosh z}dz=\int_{-\infty}^{+\infty}\frac{1}{\cosh x}dx=2\pi i\sum_k Res_{z=c_k}f(z)$

Now all we have to do is compute the residue of f in its poles. Since they're all poles of order 1 we proceed:

$\displaystyle Res_{z=c_k}f(z)=\lim_{z\rightarrow\frac{2k+1}{2}\p i i}\frac{z-\frac{2k+1}{2}\pi i}{\cosh z}=\lim_{z\rightarrow\frac{2k+1}{2}\pi i}\frac{1}{\sinh z}=\frac{1}{i\sin(\frac{2k+1}{2}\pi)}=-i(-1)^k$

so, finally we obtain:

$\displaystyle \int_{-\infty}^{+\infty}\frac{1}{\cosh x}dx=2\pi i\sum_k Res_{z=c_k}f(z)=2\pi i\sum_{k=0}^{\infty} -i(-1)^k=2\pi \sum_{k=0}^{\infty}(-1)^k$

but the series is DIVERGENT?!

Now, this means that whether I did something wrong using the residue thm., or I computed wrong the residues.

What is interesting to me is that the residue sum here appears to be actually a residue series. This is not impossible, since no residue lies ot the integraition contour and the zeroes of coshz are infinitely many, but countable.

I hope someone could help me with this puzzle

Any comments are appreciated.

Thanks a lot, Marine