7
$\begingroup$

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.

$\endgroup$
  • $\begingroup$ Hi, please see my edit of my question, where you can find the source of the problem. $\endgroup$ – Roger209 Jun 27 '15 at 9:30
  • $\begingroup$ Hello,you link can't open it,can you told us which book or which paper? $\endgroup$ – math110 Jun 27 '15 at 9:39
  • $\begingroup$ 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... $\endgroup$ – tired Jun 27 '15 at 10:29
  • $\begingroup$ @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. $\endgroup$ – Roger209 Jun 27 '15 at 12:48
  • $\begingroup$ you can further simplify parts of the integral by differentiating w.r.t. a $\endgroup$ – tired Jun 27 '15 at 12:55
2
$\begingroup$

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}

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.