How do I show that
$$
f(n,n')=\int_{0}^{\alpha}\int_{0}^{\alpha}\ln{|s-s'|}~e^{\frac{i\pi(ns+n's')}{\alpha+\beta}}\,\mathrm{d}s\,\mathrm{d}s'\\
=\begin{cases}\alpha^{2}\left(ln(\alpha)-\frac{3}{2}\right), n=n'=0\\\frac{\alpha(\alpha+\beta)}{\pi in'}\left((ln(\alpha)-1)\left(e^{\frac{\pi i\alpha n'}{\alpha+\beta}}-1\right)+\frac{\alpha+\beta}{\pi i\alpha n'}\left(e^{\frac{\pi i\alpha n'}{\alpha+\beta}}h\left(-\frac{\pi\alpha n'}{\alpha+\beta}\right)+h\left(\frac{\pi\alpha n'}{\alpha+\beta}\right)\right)\right), n=0,n'\neq0\\\frac{\alpha(\alpha+\beta)}{\pi in}\left((ln(\alpha)-1)\left(e^{\frac{\pi i\alpha n}{\alpha+\beta}}-1\right)+\frac{\alpha+\beta}{\pi i\alpha n}\left(e^{\frac{\pi i\alpha n}{\alpha+\beta}}h\left(-\frac{\pi\alpha n}{\alpha+\beta}\right)+h\left(\frac{\pi\alpha n}{\alpha+\beta}\right)\right)\right), n'=0,n\neq0\\-\frac{\alpha+\beta}{\pi n}\left(\frac{\alpha+\beta}{\pi n}ln(\alpha)e^{-\frac{\pi i\alpha n}{\alpha+\beta}}\left(e^{\frac{\pi i\alpha n}{\alpha+\beta}}-1\right)^{2}+i\alpha\left(h\left(-\frac{\pi\alpha n}{\alpha+\beta}\right)-h\left(\frac{\pi\alpha n}{\alpha+\beta}\right)\right)-\frac{\alpha+\beta}{\pi n}\left(h\left(-\frac{\pi\alpha n}{\alpha+\beta}\right)+h\left(\frac{\pi\alpha n}{\alpha+\beta}\right)+2-e^{\frac{\pi i\alpha n}{\alpha+\beta}}-e^{-\frac{\pi i\alpha n}{\alpha+\beta}}\right)\right), n+n'=0,n\neq0\\-\frac{(\alpha+\beta)^{2}}{\pi^{2}nn'}\left(ln(\alpha)\left(e^{\frac{\pi i\alpha n}{\alpha+\beta}}-1\right)\left(e^{\frac{\pi i\alpha n'}{\alpha+\beta}}-1\right)+\frac{1}{n+n'}\left(n'\left(e^{\frac{\pi i\alpha(n+n')}{\alpha+\beta}}h\left(-\frac{\pi\alpha n}{\alpha+\beta}\right)+h\left(\frac{\pi\alpha n}{\alpha+\beta}\right)\right)+n\left(e^{\frac{\pi i\alpha(n+n')}{\alpha+\beta}}h\left(-\frac{\pi\alpha n'}{\alpha+\beta}\right)+h\left(\frac{\pi\alpha n'}{\alpha+\beta}\right)\right)\right)\right), n,n',n+n'\neq0\end{cases}
$$
where $n$ and $n'$ are integers, $\alpha,\beta>0$ and
$$
h(a)=-\gamma-ln(a)+Ci(a)+Si(a)
$$
where $\gamma$ is the Euler-Mascheroni constant. The integral arises when considering surface currents on electrical conductors. Mathematica will calculate the integral analytically for individual integer values of $n$ and $n'$, in terms of the exponential integral, so I know an expression should be possible, but I'm not sure how to modify the method to this case due to the different structure of the logarithm.
Thanks in advance for any help.
Best Answer
The first integral can be decomposed into two parts to get rid of the absolute value (with $\nu=n\alpha/(\alpha+\beta)$ and $\nu'=n'\alpha/(\alpha+\beta)$): \begin{align} f(n,n')&=\int_{0}^{\alpha}\int_{0}^{\alpha}\ln{|s-s'|}e^{\frac{i(ns+n's')}{\alpha+\beta}}\,\mathrm{d}s\,\mathrm{d}s'\\ &=\alpha^2\int_{0}^{1}\int_{0}^{1}\ln{(\alpha |t-t'|)}e^{i(\nu t+\nu' t')}\,\mathrm{d}t\,\mathrm{d}t'\\ &=\alpha^2\ln\alpha \int_{0}^{1}\int_{0}^{1} e^{i\nu t+i\nu' t'}\,\mathrm{d}t\,\mathrm{d}t'+ \alpha^2 \int_{0}^{1}e^{i\nu' t'}\,dt'\int_0^{t'}e^{i\nu t}\ln(t'-t)\,dt+\\ &\hspace{3cm }+\alpha^2 \int_{0}^{1}e^{i\nu' t'}\,dt'\int_{t'}^1e^{i\nu t}\ln(t-t')\,dt\\ &=-\frac{\alpha^2\ln\alpha}{\nu\nu'}\left(e^{i\nu}-1\right)\left(e^{i\nu'}-1\right)+\alpha^2\left( I(\nu,\nu')+I(\nu',\nu) \right) \end{align} where the symmetry of the problem was used and \begin{equation} I(\nu,\nu')=\int_{0}^{1}e^{i\nu' t'}\,dt'\int_0^{t'}e^{i\nu t}\ln(t'-t)\,dt \end{equation} Integrating by parts: \begin{align} I(\nu,\nu')&= \int_0^1e^{i(\nu+\nu')t'}\,dt'\int_0^{t'}e^{-i\nu x}\ln x\,dx\\ &=-i\frac{e^{i\nu+i\nu'}}{\nu+\nu'}\int_0^1e^{-i\nu x}\ln x\,dx+i\frac{1}{\nu+\nu'}\int_0^1e^{i\nu' x}\ln x\,dx \end{align} Then, putting everything together, \begin{align} f(n,n')=-&\frac{\alpha^2\ln\alpha}{\nu\nu'}\left(e^{i\nu}-1\right)\left(e^{i\nu'}-1\right)-\\ &-i \frac{\alpha^2}{\nu+\nu'}\left[ e^{i(\nu+\nu')}\int_0^1\left( e^{-i\nu x}+e^{-i\nu' x} \right)\ln x\,dx -\int_0^1\left( e^{i\nu x}+e^{i\nu' x} \right)\ln x\,dx \right] \end{align} Defining \begin{equation} F(s)=\int_0^1e^{isx}\ln x\,dx \end{equation} \begin{equation} f(n,n')=-\frac{\alpha^2\ln\alpha}{\nu\nu'}\left(e^{i\nu}-1\right)\left(e^{i\nu'}-1\right)+i \frac{\alpha^2}{\nu+\nu'}\left[ F(\nu)+F(\nu') -e^{i(\nu+\nu')}\left( F(-\nu)+F(-\nu') \right) \right] \end{equation} Here, for real $s$, \begin{align} F(s)=&\int_0^1\cos(sx)\ln x\,dx+\int_0^1\sin(sx)\ln x\,dx\\ &=-\frac{1}{s}\int_0^1\sin(sx)\,dx-\frac{i}{s}\int_0^1\frac{1-\cos(s x)}{x}\,dx\\ &=-\frac{\operatorname{Si}(s)}s+i\frac{\operatorname{Ci}(s)-\ln(s)-\gamma}{s} \end{align} where the Euler's constant, the sine and cosine integrals are introduced. An expression in terms of the exponential integral can alternatively be obtained. I suppose that the second integral can be obtained in the same way.