# Evaluating the integral $\int_0^\infty dke^{-\gamma k}k\cos\left(\sqrt{\alpha^2k^2-\beta^2}t\right)\sin(kr)$

Consider the telegraph equation within a linear medium:

$$\left(\frac{\varepsilon\mu}{c^2}\frac{\partial^2}{\partial t^2}+\mu\sigma\frac{\partial}{\partial t}-\nabla^2\right)\Psi(\vec{r},t)=0$$

with initial conditions $\Psi(\vec{r},0)=f(\vec{r}), \Psi_t(\vec{r},0)=0$.

Taking a Fourier Transform with respect to spatial coordinates gives

$$\left(\frac{\varepsilon\mu}{c^2}\frac{\partial^2}{\partial t^2}+\mu\sigma\frac{\partial}{\partial t}+k^2\right)\hat{\Psi}(\vec{k},t)=0$$

This is a linear ODE in $t$ with characteristic equation

$$\lambda^2+\frac{\sigma c^2}{\varepsilon}\lambda+\frac{c^2k^2}{\varepsilon\mu}=0$$

Thus, assuming that $\frac{c^2k^2}{\varepsilon\mu}>\frac{\sigma^2c^4}{4\varepsilon^2}$ its solutions are given by

$$\hat{\Psi}(\vec{k},t)=e^{-\beta t}\left[A(\vec{k})\cos\left(\sqrt{\alpha^2k^2-\beta^2}t\right)+B(\vec{k})\sin\left(\sqrt{\alpha^2k^2-\beta^2}t\right)\right]$$

for arbitrary functions $A(\vec{k}), B(\vec{k})$ and $\alpha:=\frac{c}{\sqrt{\varepsilon\mu}}, \beta:=\frac{\sigma c^2}{2\varepsilon}$.

Plugging in the initial conditions, we have $A(\vec{k})=\hat{f}(\vec{k}), B(\vec{k}) = 0$, thus

$$\hat{\Psi}(\vec{k},t)=\hat{f}(\vec{k})e^{-\beta t}\cos\left(\sqrt{\alpha^2k^2-\beta^2}t\right)$$

To calculate the Inverse Fourier Transform, we calculate the Inverse Fourier Transform of $\cos\left(\sqrt{\alpha^2k^2-\beta^2}t\right)$ and then apply the Convolution Theorem.

\begin{align} \mathcal{F}^{-1}\left[\cos\left(\sqrt{\alpha^2k^2-\beta^2}t\right)\right]&=\frac{1}{8\pi^3}\int_0^\infty dkk^2\cos\left(\sqrt{\alpha^2k^2-\beta^2}t\right)\int_0^\pi d\theta\sin\theta e^{ikr\cos\theta}\int_0^{2\pi}d\phi\\ &=\frac{1}{4\pi^2}\int_0^\infty dkk^2\cos\left(\sqrt{\alpha^2k^2-\beta^2}t\right)\int_{-1}^1due^{ikru}\\ &=\frac{1}{4\pi^2r}\int_0^\infty dkk^2\cos\left(\sqrt{\alpha^2k^2-\beta^2}t\right)\left[\frac{e^{ikr}-e^{-ikr}}{ik}\right]_{u=-1}^{u=1}\\ &=\frac{1}{2\pi^2r}\int_0^\infty dkk\cos\left(\sqrt{\alpha^2k^2-\beta^2}t\right)\sin(kr) \end{align}

Since this integral does not converge, we insert a regularisation parameter and rewrite it:

$$\mathcal{F}^{-1}\left[\cos\left(\sqrt{\alpha^2k^2-\beta^2}t\right)\right]=\frac{1}{2\pi^2r}\lim\limits_{\gamma\rightarrow 0^+}\int_0^\infty dke^{-\gamma k}k\cos\left(\sqrt{\alpha^2k^2-\beta^2}t\right)\sin(kr)$$

This integral converges for all values of $\gamma>0$.

It is at this point that I am stuck.

I've tried a couple of things - writing out the product using the fact that $\cos(x)\sin(y) = \frac{1}{2}\left[\sin(x+y)-\sin(x-y)\right]$ and then rewriting this as the imaginary part of complex exponentials, using the Residue Theorem, integration by parts - in each case I get stuck because of the square root within the cosine function.

Could anybody give me a hint on how to calculate this final integral? Or does there exist no closed form?

• You are working in $\mathbb{R}^n$ for $n=$? – DisintegratingByParts Jan 16 '17 at 2:45
• $n=3$. That's where the integral for the Inverse Fourier Transform in spherical coordinates comes from. – Tom Jan 16 '17 at 2:51