7
$\begingroup$

Consider these two similar integrals

$$\int_{0}^{\infty}\sin\left({1\over 4x^2}\right)\ln x\cdot{\mathrm dx\over x^2}=-\sqrt{\pi\over 2}\cdot\left({\pi-2\gamma \over 4}\right)\tag1$$

$$\int_{0}^{\infty}\sin\left({1\over 4x^2}\right)\ln^2x\cdot{\mathrm dx\over x^2}=\sqrt{\pi\over 2}\cdot\left({\pi-2\gamma \over 4}\right)^2\tag2$$

How does one prove $(1)$ and $(2)$?

An attempt:

Using the $\sin x$ series, and let $u=\ln x$, then $(1)$ becomes

$$\sum_{n=0}^{\infty}{(-1)^n\over 2^{2n+1}(2n+1)!}\int_{-\infty}^{\infty}ue^{-2(n+1)u}\mathrm du\tag3$$

Integral $(3)$ diverges.

Another attempt:

$u={1\over 4x^2}$ then $(1)$ becomes

$$-{1\over 8}\int_{0}^{\infty}\sin u\ln{(4u)}\cdot{\mathrm du\over u^{3/2}}\tag4$$

Using $\ln u$ series, then $(4)$ becomes

$$-{1\over 4}\sum_{n=0}^{\infty}{1\over 2n+1}\int_{0}^{\infty}\left({4u-1\over 4u+1}\right)^{2n+1}\cdot{\sin u\over u^{3/2}}\mathrm du\tag5$$

Using $\coth^{-1} x={1\over 2}\ln{x-1\over x+1}$

$$-{1\over 4}\sum_{n=0}^{\infty}{1\over 2n+1}\int_{0}^{\infty}e^{2(2n+1)\coth^{-1}4x}\cdot{\sin u\over u^{3/2}}\mathrm du\tag6$$

How else can we proceed?

$\endgroup$
  • $\begingroup$ I think you may want to be careful. The series definining $\sin$ is not uniformly convergent on unbounded intervals, so you may not be allowed to integrate it term by term. $\endgroup$ – Stefano Mar 23 '17 at 12:15
  • $\begingroup$ Perhaps the keyhole contour of $\sin(1/4x^2)\ln^2(x)\frac1{x^2}$? $\endgroup$ – Simply Beautiful Art Mar 23 '17 at 13:23
  • $\begingroup$ Note that you took the series expansion of $\ln(u)$, which doesn't converge over $(0,+\infty)$... $\endgroup$ – Simply Beautiful Art Mar 23 '17 at 13:24
  • 1
    $\begingroup$ @Bui your questions are pretty fine, i'm just curious how your intuition develops $\endgroup$ – tired Mar 24 '17 at 7:55
  • 2
    $\begingroup$ This so-called intuition begin when I was 16 years, so one day I found this site accidentally, so I got a base to release all these equations and formulas off my mind. It is very stressful keeping these ideas in one mind. Maths is like a class-A drug, so addictive. $\endgroup$ – gymbvghjkgkjkhgfkl Mar 24 '17 at 8:05
8
$\begingroup$

Recall $$ \int_0^\infty u^te^{-su}du=\frac{\Gamma(t+1)}{s^{t+1}}. $$ So \begin{eqnarray} &&\int_{0}^{\infty}u^{t}\sin u\mathrm du\\ &=&\lim_{\epsilon\to0^+}\Im \int_{\epsilon}^{\infty}u^{t} e^{iu}\mathrm du=\Im \frac{\Gamma(t+1)}{(-i)^{t+1}}\\ &=&\Im \Gamma(t+1)(i)^{t+1}=\Gamma(t+1)\sin(\frac{(t+1)\pi}{2}). \end{eqnarray} Let $$ I(a)=\int_0^{\infty}x^{-a}\sin(\frac1{4x^2})dx $$ and then $$I(a)=-2^{-2+a}\int_0^{\infty}u^{\frac{-3+a}2}\sin udu=-2^{a-2}\sin(\frac{(a-1)\pi}{4})\Gamma(\frac{a-1}{2}). $$ So \begin{eqnarray} I'(2)&=&-2^{a-2}\ln2\sin(\frac{(a-1)\pi}{4})\Gamma(\frac{a-1}{2})-2^{a-4}\pi\cos(\frac{(a-1)\pi}{4})\Gamma(\frac{a-1}{2})\\ &&-2^{a-3}\sin(\frac{(a-1)\pi}{4})\Gamma'(\frac{a-1}{2})\bigg|_{a=2}\\ &=&-\ln2\cdot\frac1{\sqrt2}\sqrt{\pi}-\frac\pi 4\frac{1}{\sqrt2}\sqrt\pi-\frac{1}{2\sqrt2}\Gamma'(\frac12) \end{eqnarray} Noting $$ \frac{\Gamma'(z+1)}{\Gamma(z+1)}=-\gamma+\sum_{n=1}^\infty\left(\frac{1}{n}-\frac{1}{z+n}\right)$$ one has $$ \Gamma'(\frac12)=\Gamma(\frac12)\left[-\gamma+\sum_{n=1}^\infty\left(\frac{1}{n}-\frac{1}{n-\frac12}\right)\right]=\sqrt{\pi}(-\gamma-2\ln2).$$ Thus $$ I'(2)=-\sqrt{\pi\over 2}\left({\pi-2\gamma \over 4}\right). $$ You can treat $I''(2)$ in the same way and the calculation will be longer.

$\endgroup$

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.