How to manually infer that integral result?

Question

I would like to simplify the formula manually as follows: $$ \int_0^a{\sin \left( \frac{n\pi}{a}x \right) \frac{1}{x}\sin \left( \frac{m\pi}{a}x \right) dx} $$ Discuss the cases where (1) $n=m$ and (2) $n\neq m$, respectively.

The results were obtained by mathematica software:

Integrate[1/x*Sin[n*Pi*x/a0] Sin[n*Pi*x/a0], {x, 0, a0}] // FullSimplify

1/2 (EulerGamma - CosIntegral[2 n \[Pi]] + Log[2 n \[Pi]])

Integrate[1/x*Sin[n*Pi*x/a0] Sin[m*Pi*x/a0], {x, 0, a0}] // FullSimplify

1/2 (CosIntegral[(m - n) \[Pi]] - CosIntegral[(m + n) \[Pi]] - Log[m - n] + Log[m + n])

I.e: $$ \left\{ \begin{array}{l} \frac{1}{2}\left( \gamma -\operatorname{Ci}\left( 2n\pi \right) +\ln \left( 2n\pi \right) \right) & n=m\\ \frac{1}{2}\left( \operatorname{Ci}\left[ \left( m-n \right) \pi \right] -\operatorname{Ci}\left[ \left( m+n \right) \pi \right] +\ln \left( \frac{m+n}{m-n} \right) \right) & n\ne m\\ \end{array} \right. $$ where $\gamma$ is Euler constant, $\operatorname{Ci}$ is Cosine Integral function.

However, my ultimate goal is to know if it is possible to manually infer that result.

Answer 1

Update: For $m=n$, $$ \int_0^a\frac1{x}{\sin^2(\frac{n\pi x}{a})}dx=\int_0^a\frac{1-\cos(\frac{n\pi x}{a})}{2x}dx=\frac12(\gamma-Ci(2n\pi)+\ln(2n\pi)) $$ from here. Do the same for the other.

Answer 2

When $m \neq n$ $$ \sin \left(\frac{n \pi}{a} x\right) \sin \left(\frac{m \pi}{a} x\right) =\frac{1}{2}\left[\cos \frac{(m-n) \pi}{a} x-\cos \frac{(m+n) \pi}{a} x\right] $$ $$ \begin{aligned} I & =\frac{1}{2}\left[\int_0^a \frac{\cos \frac{(m-n) \pi x}{a} x}{x} d x-\int_0^a \frac{\cos \frac{(m+n) \pi}{a} x}{x} d x\right] \\ & =\frac{1}{2}\left[C_{in}\left(\frac{m-n}{a} \pi x\right)-C_{in}\left(\frac{m+n}{a} \pi x\right)\right]_0^a\\& =\frac{1}{2}[C_{in}((m-n) \pi)-C_{in}((m+n) \pi)] \end{aligned} $$ Using the identity in Cosine Integral,we have

$$ \begin{aligned} I & =\frac{1}{2}\left[C_i((m-n) \pi)-\gamma-\ln (m-n) \pi-C_i((m+n) \pi)+\gamma+\ln (m+n)\right]\\ & =\frac{1}{2}\left[C_i((m-n) \pi)-C_i((m+n) \pi)+\ln \left(\frac{m+n}{m-n}\right)\right] \end{aligned} $$