Double integral of a logarithm multiplied by an exponential I would like to know the value of the following integral:
$$
f(\alpha,\beta,\gamma)=\int_{0}^{1}\int_{0}^{1}\ln{(t+i\alpha t')}~e^{i(\beta t+\gamma t')}\,\mathrm{d}t\,\mathrm{d}t'
$$
where $\alpha$, $\beta$ and $\gamma$ are real-valued constants. The integral arises when considering surface currents on electrical conductors. Mathematica will calculate the integral analytically for individual integer values of the constants, in terms of the exponential integral, so I know an expression should be possible, but I'm not sure how to do this more generally. It is possible that the method in How to double integrate the product of a logarithm and an exponential may be helpful, but I don't see how to modify it to cover the difference in the argument of the logarithm here.
Thanks in advance for any help.
 A: hope you are doing well!
First things first: you are considering an integrand which is not well-behaved at $(t, t')=(0,0)$ and a complex argument for the logarithm (you must deal with branches), so you have to take care.
Start by solving the definite integral for $t$, where
\begin{equation*}
  f(\alpha, \beta, \gamma) = \int_0^1\int_0^1\log\left(t+i\alpha t'\right)e^{i(\beta t + \gamma t')} \mathrm{d}t \mathrm{d}t' = \int_0^1e^{i\gamma t'}\underbrace{\int_0^1\log\left(t+i\alpha t'\right)e^{i\beta t} \mathrm{d}t}_{:= I_1} \mathrm{d}t'.
\end{equation*}
Integrating $I_1$ by parts, see that
\begin{equation*}
 \begin{aligned}
  I_1 &= -\frac{i}{\beta}\left[e^{i\beta t}\log(t+i\alpha t')-\int_0^1\frac{e^{i\beta u}\mathrm{d}u}{u+i\alpha t'}\right]_{t=0}^{t=1} \\
  &= -\frac{i}{\beta}\left[e^{i\beta}\log(1+i\alpha t')-\log(i\alpha t') -e^{\alpha\beta t'}\int_{i\alpha t'}^{1+i\alpha t'}\frac{e^{i\beta s}\mathrm{d}s}{s}\right] \\
  &= -\frac{i}{\beta}\left[e^{i\beta}\log(1+i\alpha t')-\log(i\alpha t') -e^{\alpha\beta t'}\left(\text{Ei}(i\beta-\alpha\beta t')-\text{Ei}(-\alpha\beta t')\right)\right],
 \end{aligned}
\end{equation*}
where $\text{Ei}(x)$ is the exponential integral.
Therefore
\begin{equation*}
 \begin{aligned}
  f(\alpha, \beta, \gamma)
  &= -\frac{i}{\beta} \int_0^1e^{i\gamma t'} \left[e^{i\beta}\log(1+i\alpha t')-\log(i\alpha t') -e^{\alpha\beta t'}\left(\text{Ei}(i\beta-\alpha\beta t')-\text{Ei}(-\alpha\beta t')\right)\right] \mathrm{d}t' \\
  &= \frac{1}{\beta\gamma}\left[ e^{i\beta}\left(e^{-\frac{\gamma}{\alpha}}\text{Ei}\left(i\gamma t'+\frac{\gamma}{\alpha}\right)-e^{i\gamma t'}\log(1+i\alpha t')\right)- \text{Ei}(i\gamma t')+e^{i\gamma t'}\log(i\alpha t')\right]_{t'=0}^{t'=1} \\
  &+\frac{i}{\beta(\alpha\beta+i\gamma)}\left[\text{Ei}(i\beta-\alpha\beta t')e^{(i\gamma + \alpha\beta)t'}-e^{i\beta-\frac{\gamma}{\alpha}}\text{Ei}\left(i\gamma t'+\frac{\gamma}{\alpha}\right)\right]_{t'=0}^{t'=1} \\
  &-\frac{i}{\beta(\alpha\beta+i\gamma)}\left[\text{Ei}(-\alpha\beta t')e^{(i\gamma+\alpha\beta)t'}-\text{Ei}(i\gamma t')\right]_{t'=0}^{t'=1} \\
  &= \frac{\alpha}{\gamma(\alpha\beta+i\gamma)}\left[e^{i\beta-\frac{\gamma}{\alpha}}\left(\text{Ei}\left(i\gamma+\frac{\gamma}{\alpha}\right) - \text{Ei}\left(\frac{\gamma}{\alpha}\right)\right) - \text{Ei}(i\gamma)\right] + \frac{e^{i\gamma}}{\beta\gamma}\left(\log(i\alpha) - e^{i\beta}\log(1+i\alpha)\right)  \\
  &+\frac{\gamma_\text{EM}}{\beta\gamma}  +\frac{i}{\beta(\alpha\beta+i\gamma)}\left[(\text{Ei}(i\beta-\alpha\beta) - \text{Ei}(-\alpha\beta))e^{i\gamma+\alpha\beta} - \text{Ei}(i\beta)\right],
 \end{aligned}
\end{equation*}
where $\gamma_\text{EM}$ is the Euler-Mascheroni constant.
