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?

  • $\begingroup$ 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
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    $\begingroup$ @Bui your questions are pretty fine, i'm just curious how your intuition develops $\endgroup$ – tired Mar 24 '17 at 7:55
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    $\begingroup$ 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

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.


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