# Show $\int_{-\infty}^\infty dx \sin(ax) \log\Big(\Big| \frac{x+1}{x-1} \Big|\Big) = \frac{2 \pi \sin(a)}{a}$

I'm interested in integrals of the form $\int_{-\infty}^\infty dx f(x) \log(\big| \frac{x+1}{x-1} \big|)$. It's particularly nice when f(x) has definite parity, since $\log(\big| \frac{x+1}{x-1} \big|)$ is odd.

My question is how to show

$$I \equiv \int_{-\infty}^\infty dx \sin(ax) \log\Big(\Big| \frac{x+1}{x-1} \Big|\Big) = \frac{2 \pi \sin(a)}{a}$$ for $a>0$.

I'm trying to use complex methods to solve the integral, but I keep finding an answer of 0 owing to the lack of residues. I suspect that I'm not being careful enough about divergences and phase, and please point out if you see a mistake in the comments. Here's my work so far:

$$I = \Re \Big(\int_{-\infty}^\infty dx \sin(ax) \log\Big(\frac{x+1}{x-1} \Big) \Big)$$

I justify this line since removing the absolute value bars merely adds a finite imaginary component from the integral between $-1 < x < 1$.

Next, $$I = -\frac{1}{2}\Re \Big(i\int_{-\infty}^\infty dx\, e^{iax} \log\Big(\frac{x+1}{x-1} \Big) -i\int_{-\infty}^\infty dx\, e^{-iax} \log\Big(\frac{x+1}{x-1} \Big) \Big).$$ I believe it's kosher to split the integral into two, because each split integral converges on its own by Dirichlet's test for $|x|>1$ .

Next, I substitute $x \rightarrow -x$ in the second integral, and I find

$$I = -\frac{1}{2}\Re \Big(i\int_{-\infty}^\infty dx\, e^{iax}\, \Big(\log\Big(\frac{x+1}{x-1}\Big)-\log\Big(\frac{-x+1}{-x-1} \Big) \Big).$$

Combining the logarithms yields

$$I = -\Re \Big(i\int_{-\infty}^\infty dx\, e^{iax} \log\Big(\frac{x+1}{x-1}\Big) \Big).$$ The line above is where I wonder if some subtlety of the phase of the arguments of the logarithm was lost on me. In particular, I find that if I should have an extra '$\pi i$' added onto the logarithm, I would have the anticipated result. Continuing nonetheless, to evaluate the integral $J \equiv \int_{-\infty}^\infty dx\, e^{iax} \log\Big(\frac{x+1}{x-1}\Big)$, I choose to evaluate

$$K \equiv \oint dz\, e^{iaz} \log\Big(\frac{z+1}{z-1}\Big)$$

on the semicircle contour in the UHP, pictured below. I do not believe the branch cut I chose matters too much, since I chose the cut along the x-axis, but it could be that this choice of branch was somehow inconsistent with how I handled merging the logarithms above.

The shrinking semi-circles about -1 and 1 contribute nothing. There aren't any residues inside, so $K=0$. Thus my integral $J$ must be equal to the upper arc as $R$ goes to infinity.

Along the upper arc, note that taking the radius $R$ of the arc outward sends $\log\Big(\frac{z+1}{z-1}\Big) \rightarrow 0$ as $\frac{1}{R}$. This, coupled with Jordan's lemma, kills the upper arc as the radius goes to infinity.

Thus we find that $J=0$ and hence $I=0$, in contradiction to what we wished to show. I'll be looking over my steps to find a mistake.

Considering $$I=\int\sin(ax) \log\Big(\frac{x+1}{x-1} \Big)\,dx$$ we can integrate by parts and get $$I=-\frac 1 a \cos(ax) \log\Big(\frac{x+1}{x-1} \Big)-\frac 2 a \int \frac{ \cos (a x)}{ x^2-1}\,dx$$ The last integral is quite simple (using partial fraction decomposition and small manipulations of the trigonometric functions) $$\frac 12\Big(\cos (a)\ \text{Ci}(a(1-x))-\cos (a) \,\text{Ci}(a(1+x))+\sin (a)\, \text{Si}(a(1-x))-\sin (a)\,\text{Si}(a(1+x))\Big)$$
Now, assuming $a >0$, $$\lim_{x\to \pm\infty } \, I=\pm\frac{\pi (\sin (a)-i \cos (a))}{a}$$ and hence $$\int_{-\infty}^\infty \sin(ax) \log\Big(\frac{x+1}{x-1} \Big) \,dx=\frac{2 \pi (\sin (a)-i \cos (a))}{a}$$ $$\Re \Big(\int_{-\infty}^\infty \sin(ax) \log\Big(\frac{x+1}{x-1} \Big) \Big)\,dx=\frac{2 \pi \sin(a)}{a}$$