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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.

enter image description here

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.

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

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  • $\begingroup$ Nice work. As usual. $\endgroup$ – marty cohen Aug 26 '18 at 4:03
  • $\begingroup$ @martycohen. Please, be sure that, this coming from you, I take it as a major compliment. Thanks. $\endgroup$ – Claude Leibovici Aug 26 '18 at 4:42
  • $\begingroup$ Thanks! IBP looks like the way to go! I think you're implicitly taking a principal value in your first line when you integrate by parts, since the integral at the right isn't uniquely defined because of the singularities. However, I believe the principal value is the right way to go, since you can break up the original integral into three sections around the singularities at will and the principal value of a convergent integral is simply the convergent integral's value. Then integrating by parts yields convergent boundary terms on taking the principal value. I'll clarify further if wanted. $\endgroup$ – user196574 Aug 26 '18 at 19:11

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