What techniques that have been used to evaluate a tricky integral?
$$ \mathcal{L}G_{j,k}(\tau) = \int_0^{\infty} G_{j,k}(\tau) e^{-\tau\beta} d\tau $$ which apparently evaluates to, $$ \sum_{\mu=0}^j\sum_{\nu=0}^k \frac{ \binom{j}{\mu}\binom{k}{\nu} \left(\frac{1}{2}\right)^{j+k}(-1)^{\frac{k}{2}+\nu}B }{ (\beta\bar{\tau} +1)^2 + (\omega\bar{\tau})^2 (j+k-2 \mu -2 \nu )^2 } \quad \text{where}\ B = \cases{ \beta\bar{\tau} +1 & \text{if}\ \frac{1}{2}k \in \mathbb{N}\\ (j+k-2\mu - 2\nu)\omega\bar{\tau} & \text{if}\ \frac{1}{2}k \notin \mathbb{N} } $$
The integral is a Laplace transform of the function, $$ G_{j,k}(\tau) = re^{-r\tau} \cos ^j(\tau \omega ) \sin ^k(\tau \omega ) $$
Definitions
- $\tau$ is an exponentially distributed random variable: $P(\tau) = re^{-r\tau}$
- $\bar{\tau}$ is the average value of $\tau$
- The rate paramter, $r$, is the inverse of the average value of $\tau$ as $r = 1/\bar{\tau}$
- $\omega$ is a constant
- $j,k \in \mathbb{N}$ (assume the convention $0 \in \mathbb{N}$)
In the paper, only $(j,k) \in \{(0,0), (1,1), (2,0), (0,2)\}$ are actually used so it is possible that the Transform has only been defined in this domain. This is not explicitly indicated, however, so I assume it is general.
Current Progress
- I can easily solve for the restricted domain $(j,k) \in \{(0,0), (1,1), (2,0), (0,2)\}$ and piece together the more general form in a backwards form.
- The only method I currently know of is to solve in an iterative format and collect terms
I think the way I mentioned in the comments works well. $$\begin{align} & \int_0^\infty G_{j,k}(\tau)e^{-\tau\beta}d\tau\\ = & \int_0^\infty re^{-(\beta+r)\tau}\left(\frac{1}{2}\right)^{j+k}(e^{i\omega\tau}+e^{-i\omega\tau})^j (e^{i\omega\tau}-e^{-i\omega\tau})^k (-i)^k\\ = & \int_0^\infty re^{-(\beta+r)\tau}\left(\frac{1}{2}\right)^{j+k}(-i)^k \sum_{\mu=0}^j\sum_{\nu=0}^k (-1)^\nu e^{i\omega(2\mu+2\nu-j-k)\tau}d\tau\\ = & r\left(\frac{1}{2}\right)^{j+k}(-i)^k \sum_{\mu=0}^j\sum_{\nu=0}^k (-1)^\nu \int_0^\infty e^{(i\omega(2\mu+2\nu-j-k)-(\beta+r))\tau}d\tau\\ = & r\left(\frac{1}{2}\right)^{j+k}(-i)^k \sum_{\mu=0}^j\sum_{\nu=0}^k\frac{(-1)^\nu}{i\omega(2\mu+2\nu-j-k)-(\beta+r)} \end{align}$$ Which easily simplifies to the given equation.