# Fourier sine transform of $\frac{1}{2}+\frac{1-x^2}{4x}\ln\vert\frac{1+x}{1-x}\vert$

Show that $$\int_0^{\infty} kF(k)\sin(ka)\,dk = \frac{\pi}{2}aG(a)$$ where $$F(x) = \frac{1}{2}+\frac{1-x^2}{4x}\ln\vert\frac{1+x}{1-x}\vert$$ and $$G(x) = \frac{\sin x-x\cos x}{x^4}$$

EDIT: The source can be found here. One should notice that the function $F(x)$ is not continuous at $x=1$.

EDIT2: The integral below may be of some help. $$\ln\vert\frac{a+x}{a-x}\vert = 2\int_0^{\infty}\frac{\sin at\sin xt}{t}\,dt$$

EDIT3: Here is a mathematica code of my question:

F[x_] := (1/2 + (1 - x^2)/(4 x)*Log[Abs[(1 + x)/(1 - x)]])xSin[a*x];

Integrate[F[x], {x, 0, Infinity}, Assumptions -> a > 0]

The result is:

([Pi] (-a Cos[a] + Sin[a]))/(2 a^3)

EDIT4: By virtue of Bessel function $J_{\frac{1}{2}}(z)=\sqrt{\frac{2}{\pi z}}\sin(z)$, the integral identity turns to be $$\int_0^{\infty} k^{\frac{3}{2}}F(k)J_{\frac{1}{2}}(ka)\,dk = \sqrt{\frac{\pi}{2}a}G(a)$$ In this occasion, some refs are useful:

(1) Lin Q.G.: Inﬁnite integrals involving Bessel functions by an improved approach of contour integration and the residue theorem.

(2) Lucas S.K.: Evaluating inﬁnite integrals involving Bessel functions of arbitrary order.

• Hi, please see my edit of my question, where you can find the source of the problem. – Roger209 Jun 27 '15 at 9:30
• Hello,you link can't open it,can you told us which book or which paper? – math110 Jun 27 '15 at 9:39
• i think this may be done by using a suitable integral representation of the logarithmic term. Maybe one that is obtained by a Frullani like integral. thinking about it... – tired Jun 27 '15 at 10:29
• @math110: Sorry for the disable link, I have corrected it. In fact, this integral relates to the well-known phenomenon in solid-state theory, the so-called RKKY interaction. – Roger209 Jun 27 '15 at 12:48
• you can further simplify parts of the integral by differentiating w.r.t. a – tired Jun 27 '15 at 12:55

The integral can be written as \begin{align} I&=\frac{1}{2}\int_0^{\infty} k\sin(ka)\left[1+\frac{1-k^2}{2k}\ln\left|\frac{1+k}{1-k}\right|\right]\,dk\\ &=\frac{1}{2}\int_0^{\infty} \left( 1-k^2 \right)\sin(ka)\left[\frac k{1-k^2}+\frac{1}{2}\ln\left|\frac{1+k}{1-k}\right|\right]\,dk\\ &=\frac{1}{2}\left( 1+\frac{d^2}{da^2} \right)\int_0^{\infty} \sin(ka)\left[\frac k{1-k^2}+\frac{1}{2}\ln\left|\frac{1+k}{1-k}\right|\right]\,dk \end{align} Now, recognizing that \begin{equation} \frac k{1-k^2}+\frac{1}{2}\ln\left|\frac{1+k}{1-k}\right|=\frac{1}{2}\frac{d}{dk}\left( k\ln\left|\frac{1+k}{1-k}\right| -2\right) \end{equation} The function $g(k)= k\ln\left|\frac{1+k}{1-k}\right| -2$ is is such that $g'(k)$ is integrable, $g(k)\sim 2k^{-2}/3$ for $k\to\infty$ and $g(k)\sim -2$ for $k\to 0$. It comes \begin{align} I&=\frac{1}{4}\left( 1+\frac{d^2}{da^2} \right)\int_0^{\infty} \sin(ka)\frac{d}{dk}\left( k\ln\left|\frac{1+k}{1-k}\right| -2\right)\,dk\\ &=-\frac{1}{4}\left( 1+\frac{d^2}{da^2} \right)a\int_0^{\infty} \cos(ka)\left[k\ln\left|\frac{1+k}{1-k}\right| -2\right]\,dk\\ &=-\frac{1}{4}\left( 1+\frac{d^2}{da^2} \right)a\frac{d}{da}\int_0^{\infty} \sin(ka)\left[\ln\left|\frac{1+k}{1-k}\right|-\frac{2}{k}\right] \,dk \end{align} We know (GR 17.33.35) that \begin{equation} \ln\left|\frac{1+k}{1-k}\right|=2\int_0^\infty\sin t\sin kt \frac{dt}{t} \end{equation} and \begin{equation} \int_0^{\infty} \frac{2\sin(ka)}{k}dk=\pi \end{equation} thus \begin{equation} I=-\frac{1}{2}\left( 1+\frac{d^2}{da^2} \right)a\frac{d}{da}\left[\int_0^{\infty} \sin(ka)\,dk\int_0^\infty\sin t\sin kt \frac{dt}{t}-\pi\right] \end{equation} The contribution of the constant term in the bracket vanishes as it does not depend on $a$. The double integral is the sine transform of the sine transform of $\sin t/t$. Finally, \begin{align} I&=-\frac{\pi}{4}\left( 1+\frac{d^2}{da^2} \right)a\frac{d}{da}\left( \frac{\sin a}{a} \right)\\ &=\frac{\pi}{2}\frac{\sin a -a\cos a}{a^3} \end{align}