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I have been asked to evaluate $$\int_{-\infty}^\infty \frac{dx}{\cosh(x-a)\cos(2x)}$$. I'm deliberating on whether this indefinite integral exists or not. The integrand diverges when $x=\frac{1}{2}(n+\frac{1}{2})\pi$ but the $\cosh(x-a)$ term relaxes these singularities exponentially as $x$ goes to infinity.

If it does exist, then I'm left with the problem of computing it. This is for a complex variables class, so I was using residue methods. However, there are countably many simple poles along the real axis as well as for each $z=i(n+\frac{1}{2})\pi +a$ in the complex plane, so I don't even know what contour to use. It looks like rectangles and semicircles are out. Any suggestions?

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    $\begingroup$ This integral looks a bit intimidating to me. But I am just wondering. What happens if x = pi/4 ? That cosine term is buggin me... $\endgroup$ – imranfat Oct 22 '13 at 19:36
  • $\begingroup$ The Cauchy principal value of the integral under consideration makes sense. However, both Maple and Mathematica fail with it. $\endgroup$ – user64494 Oct 22 '13 at 19:56
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My calculation shows that

$$ \mathrm{PV}\!\!\int_{-\infty}^{\infty} \frac{dx}{\cosh(x-a)\cos x} = 2\pi \sum_{k=0}^{\infty} \frac{(-1)^{k}}{\cosh \left( \pi k + \frac{\pi}{2} \right)} \cos(2k+1)a. \tag{1} $$

Indeed, what I derived is that

$$ \int_{-\infty}^{\infty} \frac{dx}{\cosh(x-a)\cos (x+i\epsilon)} = 2\pi \sum_{k=0}^{\infty} \frac{(-1)^{k}e^{-(2k+1)\epsilon}}{\cosh \left( \pi k + \frac{\pi}{2} \right)} e^{i(2k+1)a}, \quad \epsilon > 0. $$

I have no idea how to simplify $\text{(1)}$, except for the following obvious special case:

$$ \mathrm{PV}\!\!\int_{-\infty}^{\infty} \frac{dx}{\cosh\left(x-\frac{\pi}{2}\right)\cos x} = \mathrm{PV}\!\!\int_{-\infty}^{\infty} \frac{dx}{\cosh x \sin x} = 0. $$

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So it turns out that the integral is definitely not convergent (as expected). Consider a small interval about $2x=\frac{\pi}{2}$, ie $x=\frac{\pi}{4}$. Then $\frac{1}{\cos(2x)}$ behaves like $\frac{1}{x}$ in this small interval. Also the $\cosh(x-a)$ term is non-zero so has little influence; it can be treated like a constant. But $\int \frac{1}{x} dx = \log x$, which diverges in this small interval so the integral diverges over the real line also.

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