3

I can't seem to find this problem on Math.SE, so sorry if it actually exists.

I want to evaluate $I(a) = \int_0^{\infty} \frac{\log(ax)\sin x}{x} dx$. Then I separated it into two integrals $I(a) = \int_0^{\infty} \frac{\log(a)\sin x}{x} + \int_0^{\infty} \frac{\log(x)\sin x}{x}$. The first term is constant and evaluates to $\frac{\pi}{2}\log a$, and the second term is where I am struggling with, it's equal to $I(1)$.

I also tried using differentiation under the integral sign, but it gives the same result, as I get $I'(a) = \frac{\pi}{2a}$, and I can't find the constant.

I also tried putting the variable inside $\sin(x)$, so $J(a) = \int_0^{\infty} \frac{\log x \sin(ax)}{x}$, but $I'(a) = \int_0^{\infty} \log x \cos(ax) dx$ is divergent.

WolframAlpha gives something about $\pi\gamma$, so I am wondering where that comes from.

Thanks!

Leucippus
  • 27,174
ketsi
  • 3,904
  • Try here : https://math.stackexchange.com/questions/1941888/calculate-int-0-infty-frac-sinx-logxx-mathrm-dx?noredirect=1 or https://math.stackexchange.com/questions/3053019/integral-with-euler-mascheroni-constant?noredirect=1. – HackR Nov 15 '22 at 02:36
  • If you are looking for a hint, try looking at Laplace transforms for a quick method to solve this. – HackR Nov 15 '22 at 02:37
  • $$I=\int_0^{\infty}\frac{\log(x)\sin(x)}{x} dx=\frac{\partial}{\partial \epsilon}\bigg|{\epsilon=0}\Im\int_0^{\infty} e^{ix}x^{1-\epsilon}dx=\frac{\partial}{\partial \epsilon}\bigg|{\epsilon=0}\Im I_\epsilon$$ Making a turn in the complex plane (what is equivalent to the change $x=te^\frac{\pi i}{2}$; integral along a quarter-circle $\to 0$ as the radius $\to\infty$) $$I_\epsilon=e^\frac{\pi i\epsilon}{2}\int_0^\infty e^{-t}t^{\epsilon-1}dt,\Rightarrow, \Im I_\epsilon =\sin\frac{\pi \epsilon}{2}\Gamma(\epsilon)=\frac{\sin\frac{\pi \epsilon}{2}}{\epsilon}\Gamma(1+\epsilon)$$ – Svyatoslav Nov 15 '22 at 02:42
  • $$\Gamma(1+\epsilon)=1-\gamma\epsilon+O(\epsilon^2); \quad\frac{\sin\frac{\pi\epsilon}{2}}{\epsilon}=\frac{\pi}{2}+O(\epsilon^2)$$ – Svyatoslav Nov 15 '22 at 02:46
  • 1
    Thank you for your help. @Svyatoslav I am not too familiar with complex analysis, though I will get some paper and see if the math makes sense haha. @ HackR I don't know laplace transformation. This is from an integration competition (essentially integration bee...) and laplace transform is out of syllabus too – ketsi Nov 15 '22 at 04:13

1 Answers1

3

By using robjohn's answer Integral then one can determine that $$ \int_{0}^{\infty} \frac{\ln(x) \, \sin(b x)}{x} \, dx = - \frac{\pi}{2} \, \text{sgn}(b) \, (\ln|b| - \gamma). $$ As seen by the basis of the proposed problem the integral $$ \int_{0}^{\infty} \frac{\ln(a x) \, \sin(b x)}{x} \, dx $$ then follows as: \begin{align} I &= \int_{0}^{\infty} \frac{\ln(a x) \, \sin(b x)}{x} \, dx \\ &= \ln(a) \, \int_{0}^{\infty}\frac{\sin(b x)}{x} \, dx + \int_{0}^{\infty} \frac{\ln(x) \, \sin(b x)}{x} \, dx \\ &= \ln(a) \, \frac{\pi}{2} \, \text{sgn}(b) - \frac{\pi}{2} \, \text{sgn}(b) \, ( \ln|b| + \gamma) \\ &= \frac{\pi}{2} \, \text{sgn}(b) \, \ln\left(\frac{a}{|b|} \, e^{-\gamma}\right). \end{align} Here, $\text{sgn}(x)$ is the sign of $x$ and is $+1$ for $x>0$ and $-1$ otherwise.

Leucippus
  • 27,174