I came across the following integral:

$$\int_0^{\infty} e^{-x^2} \frac{\sin(a x)}{\sin(b x)} dx$$

while trying to calculate the inverse Laplace transform

$$ L_p^{-1} \left[ \frac{\sinh(\alpha\sqrt{p})}{\sinh(\beta\sqrt{p})} \frac{e^{-\gamma\sqrt{p}}}{\sqrt{p}} \right], |\alpha|<\beta, \gamma>0$$

using the Bromwich integral approach. The contour I used is the following:

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the above mentioned integral arises while doing integration over the segments $L_1^+,L_2^+,\cdots$ and $L_1^-,L_2^-,\cdots$.

I have searched for this integral in Prudnikov et. al., Integrals and Series, v.1, but found nothing. I have also tried to evaluate the integral using residue theorem, but could not quite decide which contour to use.

Any help is greatly appreciated!

P.S. The ILT can be calculated by noticing that $$ F[p] = \frac{\sinh (\sqrt{p} \alpha)}{\sinh (\sqrt{p} \beta)} \frac{e^{-\gamma\sqrt{p}}}{\sqrt{p}} = \sum_{n=0}^{\infty} \left( \frac{e^{-(-\alpha+\beta+\gamma+2n\beta)\sqrt{p}}}{\sqrt{p}} -\frac{e^{-(\alpha+\beta+\gamma+2n\beta)\sqrt{p}}}{\sqrt{p}} \right)$$ using $$L_p^{-1} \left[ \frac{e^{-\alpha\sqrt{p}}}{\sqrt{p}} \right] = \frac{1}{\sqrt{\pi t}} e^{-\frac{\alpha^2}{4t}}$$ we get $$\begin{align*} f(t) &= L_p^{-1}[F(p)] \\ &= \sum_{n=0}^{\infty} \left( \frac{ e^{-(-\alpha+\beta+\gamma+2n\beta)^2/4t} }{\sqrt{\pi t}} - \frac{ e^{-(-\alpha+\beta+\gamma+2n\beta)^2/4t} }{\sqrt{\pi t}} \right). \end{align*}$$ Here I am more interested in calculating the above ILT using the Bromwich integral approach.


2 Answers 2


The actual integral I get after taking the ILT is

$$PV \int_0^{\infty} dx \, \frac{\sin{\alpha x}}{\sin{\beta x}} \cos{\gamma x} \, e^{-t x^2} $$

where $PV$ denotes the Cauchy principal value of the integral. Note that we must take this $PV$ of this integral as there are poles in the integrand - without it, the integral is infinite. I guess generally speaking this integral is equivalent to a sum of two of the integrals you specify - except that $\alpha$ and $\beta$ take on different meanings. I do not know how to evaluate this integral directly.

Note that, in taking the ILT, you must use a modified Bromwich contour that avoids the branch cut along the negative real axis. Unfortunately, as you have poles on that axis, you need to provide semicircular detours on each traversal above and below the branch cut at each pole. Basically, you replace $-1=e^{i \pi}$ above the branch cut and $-1=e^{-i \pi}$ below. That way, you end up with, as the ILT, an expression as follows:

$$\begin{align}\frac1{i 2 \pi} \int_{c-i \infty}^{c+i \infty} dp \, \frac{\sinh{\alpha \sqrt{p}}}{\sinh{\beta \sqrt{p}}} \frac{e^{-\gamma \sqrt{p}}}{\sqrt{p}} &= \frac1{\pi} PV \int_0^{\infty} dx \, \frac{\sin{\alpha x}}{\sin{\beta x}} \cos{\gamma x} \, e^{-t x^2} \\&+ \frac1{\beta} \sum_{n=1}^{\infty} (-1)^n \sin{\left (\frac{\alpha}{\beta} n \pi \right )} \sin{\left (\frac{\gamma}{\beta} n \pi \right )} e^{-n^2 \pi^2 t/\beta^2}\end{align}$$

Again, I have do not know how to evaluate the integral or sum in closed form in general.

  • $\begingroup$ Thanks Ron. This is exactly what I have got. The infinite series part is not a problem as it has a nice convergence property, and many ILT of similar functions involving sinh and cosh are expressed in infinite series (or theta function, see e.g. Prudnikov et. al., Integrals and Series, v.5, 2.3.2; Erdelyi, et.al., Tables of Integral Transforms, 5.9(32)). The integral part is the one that bothers me. $\endgroup$ Mar 1, 2016 at 4:03
  • $\begingroup$ @PengchengSong: I love the contour diagram! Little wonder we got the same result. I am sorry to say, I do not know how to evaluate the integral unless, e.g., $\alpha = \gamma = \beta/2$. $\endgroup$
    – Ron Gordon
    Mar 1, 2016 at 9:26

Given $$ L_p^{-1} \left[ \frac{\sinh(\alpha\sqrt{p})}{\sinh(\beta\sqrt{p})} \frac{e^{-\gamma\sqrt{p}}}{\sqrt{p}} \right], |\alpha|<\beta, \gamma>0$$ this amounts to finding the poles involved of the function. In this case what needs to be found are the poles of $\sqrt{p}=0$ and $\sinh(\beta \sqrt{p})=0$. In both cases $p=0$ is a pole and by using $\sinh(x) = - i \, \sin(i x)$ then $\sin(i \beta \sqrt{p}) = 0$ leads to $p_{n} = - (n^2 \, \pi^2)/\beta^2$, for $n \geq 0$.

For the case of $p_{n}$ it is seen that: \begin{align} \lim_{p \to p_{n}} \left\{ (p - p_{n}) \, \frac{\sinh(\alpha\sqrt{p})}{\sinh(\beta\sqrt{p})} \frac{e^{-\gamma\sqrt{p}}}{\sqrt{p}} \, e^{p \, t} \right\} &= \frac{\sinh(\alpha \sqrt{p_{n}})}{\sqrt{p_{n}}} \, e^{p_{n} \, t - \gamma \, \sqrt{p_{n}}} \, \lim_{p \to p_{n}} \left\{ \frac{p - p_{n}}{\sinh(\beta \sqrt{p})} \right\} \\ &= \frac{\sinh(\alpha \sqrt{p_{n}})}{\sqrt{p_{n}}} \, e^{p_{n} \, t - \gamma \, \sqrt{p_{n}}} \, \lim_{p \to p_{n}} \left\{ \frac{1}{\cosh(\beta \sqrt{p})} \right\} \\ &= \frac{2 \, \sinh(\alpha \sqrt{p_{n}})}{\beta \, \cosh(\beta \sqrt{p_{n}})} \, e^{p_{n} \, t - \gamma \, \sqrt{p_{n}}} \\ &= \frac{2 i}{\beta} \, (-1)^{n} \, \sin\left(\frac{n \pi \, \alpha}{\beta}\right) \, e^{- \frac{n^2 \pi^2 \, t}{\beta^2} - i \, \frac{\gamma \, n \pi}{\beta}}. \end{align} A derivative and limit will take place for the pole of zero, since it is of order two. Once that value is found then it follows that $$L^{-1}\{f(s)\} = 2\pi i \sum \{Res\}$$

  • 3
    $\begingroup$ This completely ignores the branch cut along the negative axis. $\endgroup$
    – Ron Gordon
    Mar 1, 2016 at 3:12

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