# Prove $\int^{\infty}_0 \frac{\sin(\sqrt{x})\ln^2\big(\frac{1}{\sqrt{x}}\big)}{x}dx=\frac{\pi^3}{12}+\gamma^2 \pi$

I came across the following integral

Prove $$\int^{\infty}_0 \frac{\sin(\sqrt{x})\ln^2\big(\frac{1}{\sqrt{x}}\big)}{x}dx=\frac{\pi^3}{12}+\gamma^2 \pi$$

where $$\gamma$$ is the euler-mascheroni constant. I simplify the integral. let $$x=u^2 \Rightarrow dx=2udu$$

$$\Rightarrow\int^{\infty}_0 \frac{\sin(\sqrt{x})\ln^2\big(\frac{1}{\sqrt{x}}\big)}{x}dx\ = 2\int^{\infty}_0 \frac{\sin(u)\ln^2(u)}{u}du$$

How does one proceed from here? Is there a particular formula or theorem that I need to use? Am I in too deep water if I can't figure this out?

• what is $\gamma$? is it the euler-mascheroni constant? Sep 5, 2021 at 6:28
• @IshraaqParvez Yes it is I have edited the post to clarify Sep 5, 2021 at 6:31
• Shouldn't that be $ln^2(\frac{1}{u})$? Sep 5, 2021 at 6:46
• @BrianVandenberg $\ln^2(1/u))=(-\ln(u))^2=\ln^2(u)$ Sep 5, 2021 at 6:50
• Try laplace transform. Sep 5, 2021 at 6:58

We start with the substitution $$x\mapsto x^2$$ and noting that $$\ln^2(1/x) = \ln^2x$$.
$$I =2 \int_0^\infty \frac{\sin(x)\ln^2x}{x}\, dx$$ Now, let $$I(k) = \int_0^\infty \frac{\sin x}{x^k}\, dx\qquad 0\lt k\le 1$$ Also, we have $$\frac1{x^k} =\frac1{\Gamma(k)} \int_0^\infty e^{-xt}t^{k-1}\, dt$$ Using this result \begin{align} I (k)&= \frac1{\Gamma(k)} \int_0^\infty\int_0^\infty e^{-xt}\sin(x)t^{k-1}\, dtdx \\ &= \frac1{\Gamma(k)} \int_0^\infty\int_0^\infty e^{-xt}\sin(x)t^{k-1}\, dxdt \\ &= \frac1{\Gamma(k)}\int_0^\infty \frac{t^{k-1}}{t^2+1} \, dt \\ I(k) &= \frac{\pi}{2\Gamma(k)\sin\frac{\pi k}{2}} \end{align} To get the last result, we use the substitution $$t^2 \mapsto t$$ and the classical result $$\int_0^\infty \frac{t^{s-1}}{1+t}\, dt = \frac{\pi}{\sin (\pi s)}$$. I wasn't able to justify the interchange of the two integrals.
Now differentiating both sides twice w.r.t. $$k$$ and letting $$k \to 1$$ we get $$\int_0^\infty \frac{\sin(x)\ln^2x}{x}\, dx= \frac I2 = \frac{\pi \gamma^2}{2}+\frac{\pi^3}{24}$$