I need help solving the following integral:
$$ \int_0^\infty dx \frac{e^{2ikax}}{\cosh^2(x)} $$
I used integration by parts (as well as $\int_0^\infty dx \frac{1}{\cosh^2(x)}=1$, which I looked up in a table of integrals) as follows:
$$\int_0^\infty dx \frac{e^{2ikax}}{\cosh^2(x)}=\Big[1 \cdot e^{2ikax} \Big]_0^\infty-\int_0^\infty dx \cdot 1 \cdot 2ika \cdot e^{2ikax}$$
However the $\Big[1 \cdot e^{2ikax} \Big]_0^\infty$ part diverges, though according to my excercise sheet the integral has a non-zero solution.
Can anyone provide me a tip or clue as to what I am doing wrong? Thanks!
You've made an $\infty - \infty$ error; you need to refine your analysis.
Recall that the integral to infinity is defined as
$$ \int_0^\infty f(x) \, \mathrm{d}x = \lim_{B \to \infty} \int_0^B f(x) \, \mathrm{d} x $$
You can apply your approach via integration by parts to the integral within the limit.
Your aim in doing so is generally to take enough care to cancel out the parts contributing the infinite values in the difference; once you've done so, you can evaluate to simplify the other parts and continue with the calculation.
Incidentally, you've made an error in your integration by parts: presumably you set
$$ \begin{align} u = e^{2ikax} &\qquad \qquad& \mathrm{d}v = \frac{\mathrm{d}x}{\cosh(x)^2} \\ \mathrm{d}u = 2ika e^{2ikax}\mathrm{d}x &\qquad \qquad& v = \int_0^{\infty} \frac{\mathrm{d}x}{\cosh(x)^2} = 1 \end{align}$$
However, your calculation of $v$ is incorrect! You need the antiderivative, not a particular definite integral of $v$. In particular, it's clear that $\mathrm{d}1 \neq \mathrm{d}v$
Instead, what you need is $$ \mathrm{d}u = 2ika e^{2ikax} \mathrm{d}x\qquad \qquad v = \int \frac{\mathrm{d}x}{\cosh(x)^2} = \tanh(x) $$
(any antiderivative will do, so I can pick the constant of integration to be whatever I like)
So, the correct formula is
$$ \lim_{B \to \infty} \left( \left[ e^{2ikax} \tanh(x) \right]_0^B - \int_0^B \tanh(x)2ika e^{2ikax} \, \mathrm{d}x \right) $$