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Here is my approach

$$\int_0^\infty \frac{\cos(\alpha x)}{(x^2 + {\beta}^2)\cosh(\pi x)} \, dx$$

Using contour integration and residue calculus, the poles at $x = \pm i\beta$ and $x = i(n + 1/2)$ for integers $n$.

Residue at $x = i{\beta}$: $$ \text{Res}_{x = i{\beta}} \frac{e^{i\alpha x}}{(x^2 + {\beta}^2)\cosh(\pi x)} = \frac{e^{-\alpha {\beta}}}{2i{\beta} \cos(\pi {\beta})} $$

Residues at $x = i(n + 1/2)$: $$ \text{Res}_{x = i(n + 1/2)} \frac{e^{i\alpha x}}{(x^2 + {\beta}^2)\cosh(\pi x)} = \frac{i (-1)^{n + 1} e^{-\alpha(n + 1/2)}}{ \pi ({\beta}^2 - (n + 1/2)^2) } $$

Summing these residues and multiplying by $2\pi i$, we obtain the integral over the entire real line.

$$\frac{\pi e^{-\alpha {\beta}}}{2{\beta} \cos(\pi {\beta})} + \sum_{n=0}^\infty \frac{(-1)^{n+1} e^{-\alpha(n + 1/2)}}{(n + 1/2)^2 - {\beta}^2} $$

equalityisillusion
  • 129

1 Answers1

6

Applying partial fraction decomposition to the summand and invoking Lerch transcendent, OP's formula reduces to:

$$ \begin{align*} I(\alpha, \beta) &= \frac{\pi e^{-\alpha\beta}}{2\beta \cos(\pi\beta)} + \frac{e^{-\alpha/2}}{2\beta}[\Phi(-e^{-\alpha}, 1, \tfrac{1}{2}+\beta) - \Phi(-e^{-\alpha}, 1, \tfrac{1}{2}-\beta)]. \end{align*} $$

As to my knowledge, this expression cannot be further simplified in terms of more elementary functions (without invkoing other transcendental functions).

Addendum. I was able to replicate OP's formula using an alternative approach. Here, we will assume $\alpha \geq 0$ and $\beta > 0$ WLOG. Using the identity

$$ \frac{1}{\cosh(\pi \xi)} = \int_{-\infty}^{\infty} \frac{e^{2\pi i \xi x}}{\cosh(\pi x)} \, \mathrm{d}x, $$

we can give a convolution-type formula:

$$ \begin{align*} I(\alpha, \beta) &= \frac{1}{4\beta} \int_{-\infty}^{\infty} \frac{e^{-\beta|\alpha-x|}}{\cosh(x/2)} \, \mathrm{d}x. \end{align*} $$

This strongly suggests that we cannot avoid using transcendental functions for finding a closed-form of $I$. Including them into our vocabulary of closed forms, we can proceed as follows:

$$ \begin{align*} I(\alpha, \beta) &= \frac{e^{-\alpha\beta}}{2\beta} \int_{0}^{\infty} \frac{\cosh(\beta x)}{\cosh(x/2)} \, \mathrm{d}x + \frac{1}{4\beta} \int_{\alpha}^{\infty} \frac{e^{\alpha\beta} e^{-\beta x} - e^{-\alpha\beta} e^{\beta x}}{\cosh(x/2)} \, \mathrm{d}x \\ &= \frac{\pi e^{-\alpha\beta}}{2\beta \cos(\pi\beta)} + \frac{1}{4\beta} \int_{\alpha}^{\infty} \frac{e^{\alpha\beta} e^{-\beta x} - e^{-\alpha\beta} e^{\beta x}}{\cosh(x/2)} \, \mathrm{d}x \end{align*} $$

To further simplify, we introduce an auxiliary function:

$$ J(\alpha, \beta) \triangleq \frac{e^{\alpha\beta}}{2} \int_{\alpha}^{\infty} \frac{e^{-\beta x}}{\cosh(x/2)} \, \mathrm{d}x. $$

Then

$$ \begin{align*} J(\alpha, \beta) &= e^{\alpha\beta} \sum_{n=0}^{\infty} (-1)^n \int_{\alpha}^{\infty} e^{-(\beta+n+\frac{1}{2}) x} \, \mathrm{d}x \\ &= e^{-\alpha/2} \sum_{n=0}^{\infty} \frac{(-1)^n e^{-n \alpha}}{\beta+n+\frac{1}{2}} \\ &= e^{-\alpha/2} \Phi(-e^{-\alpha}, 1, \beta+\tfrac{1}{2}), \end{align*} $$

where $\Phi$ is the Lerch transcendent. Now using this, $I(\alpha, \beta)$ can be written as:

$$ \begin{align*} I(\alpha, \beta) &= \frac{\pi e^{-\alpha\beta}}{2\beta \cos(\pi\beta)} + \frac{J(\alpha, \beta) - J(\alpha, -\beta)}{2\beta} \\ &= \frac{\pi e^{-\alpha\beta}}{2\beta \cos(\pi\beta)} + \frac{e^{-\alpha/2}}{2\beta}[\Phi(-e^{-\alpha}, 1, \tfrac{1}{2}+\beta) - \Phi(-e^{-\alpha}, 1, \tfrac{1}{2}-\beta)]. \end{align*} $$

Sangchul Lee
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