Evaluate integral $\int_0^1\sin(\pi \sqrt{-\ln x})\,dx$

Question

$$I=\int_0^1\sin(\pi \sqrt{-\ln x})\mathrm~ dx$$

My efforts:

I substitute $t=-\ln x$

And I got $$I=\int_0^\infty e^{-t}\sin(\pi \sqrt{t})\mathrm ~dt$$

But, how to proceed with : $\int_0^\infty e^{-t}\sin(\pi\sqrt{t})\mathrm dt$?

(P.S.: The result is $\dfrac{\pi^{3/2}}{2e^{\pi^2/4}}$)

Welcome to MSE. Please, use MathJax to type the math in your questions instead of using unusual characters that won't render properly in many devices. Consider also that many users in this site won't even read your question if it's not properly formatted, let alone help you. — jjagmath, Jul 07 '23 at 17:25
How did you get $e^{-t}\sin(π \sqrt{t})$? The substitution $t = -\ln x$ gives $dt=-1/x dx$. — cpiegore, Jul 07 '23 at 17:55
You ask one question in the title, answer it, and then post a different question as the "real" one. Don't you see any awkwardness in that? — David G. Stork, Jul 07 '23 at 17:59
There is some elegant mathematics going on here, but unfortunately the question was a PSQ and got closed (and is unlikely to be reopened given its vote score). Would it be worthwhile to repurpose this as a Q&A? — Oscar Lanzi, Jul 07 '23 at 18:29
Duplicate: Integrate $ x\cdot e^{i\omega x - x^2}$ from $0$ to $\infty$ — Anne Bauval, Jul 08 '23 at 14:45

score 0 · Answer 1 · answered Jul 08 '23 at 19:23

Let $t=\sqrt{-\ln x}$, hence $x=e^{-t^2}$. Integration by part, and we get

$$I=\int_\infty^0\sin(\pi t)~d(e^{-t^2})=\sin(\pi t)~d(e^{-t^2})\bigg|_\infty^0-\pi\int_\infty^0\cos(\pi t)\cdot e^{-t^2}dt=\pi\int^\infty_0\cos(\pi t)\cdot e^{-t^2}dt$$

This is a Gaussian-like integral and can be solved by many ways

$$\int_0^\infty \cos(ax)\cdot e^{-x^2}~dx=\frac{\sqrt\pi}{2}\,e^{-a^2/4}$$

therefore,

$$\boxed{\int_0^1\sin(\pi \sqrt{-\ln x})\mathrm~ dx=\frac{\pi\sqrt\pi}{2}\,e^{-\pi^2/4}}$$

Evaluate integral $\int_0^1\sin(\pi \sqrt{-\ln x})\,dx$

1 Answers1