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

Question

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 $a>0$.

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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.

Answer 1

Without contour integration.

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 $$\int_{-\infty}^\infty \sin(ax) \log\Big(\frac{x+1}{x-1} \Big) \,dx=\frac{2 \pi (\sin (a)-i \cos (a))}{a}$$ $$\Re \Big(\int_{-\infty}^\infty \sin(ax) \log\Big(\frac{x+1}{x-1} \Big) \Big)\,dx=\frac{2 \pi \sin(a)}{a}$$

Answer 2

If we use the principal branch of the logarithm, then $\log \left(\frac{z+1}{z-1} \right)$ has a branch cut along the real axis from $-1$ to $1$ (since that is where $\frac{z+1}{z-1}$ is real and negative).

On the upper side of the branch cut, the argument of $\frac{z+1}{z-1}$ is $-\pi$. (See below.)

So if we integrate the function $$e^{iaz} \log \left(\frac{z+1}{z-1} \right) \ , \quad a>0, $$ around the contour depicted in your question, we end up with

$$ \small\int_{-\infty}^{-1} e^{iax} \log \left(\frac{x+1}{x-1} \right) \, \mathrm dx + \int_{-1}^{1} e^{iax}\left( \log \left(\left|\frac{x+1}{x-1} \right|\right) - \pi i\right) \mathrm dx + \int_{1}^{\infty}e^{iax} \log \left(\frac{x+1}{x-1} \right) \, \mathrm dx = 0 \, , $$

from which we get

$$\int_{-\infty}^{\infty} e^{iax} \log \left(\left|\frac{x+1}{x-1} \right|\right) \, \mathrm dx = \pi i \int_{-1}^{1} e^{iax} \, \mathrm dx = \frac{\pi}{a} \left(e^{ia}-e^{-ia} \right) = \frac{2 \pi i \sin(a)}{a}.$$

Now just equate the imaginary parts on both sides of the equation.

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To see that the argument of $\frac{z+1}{z-1}$ on the upper side of the branch cut is $-\pi$ and not $\pi$, notice that $$\frac{z+1}{z-1} = \frac{x+iy+1}{x+iy-1} = \frac{x^{2}+y^{2}-1}{(x-1)^{2}+y^{2}} - \frac{2iy}{(x-1)^{2}+y^{2}}.$$

Therefore, for all points in the upper half-plane, $- \pi < \operatorname{Arg} \left(\frac{z+1}{z-1} \right) < 0.$

And for all points in the lower half-plane, $0 < \operatorname{Arg} \left(\frac{z+1}{z-1} \right) < \pi. $