I have stumbled onto an interesting integral$$\int_0^\infty \sin(x)\sin\left(\frac{1}{x}\right)dx$$ which I noticed graphically that it appears to be $1$, but I have no idea on how to evaluate it. Maybe it could be done with the use of Bessel Functions? Any help is appreciated.
Closed form of $\int_0^\infty \sin(x)\sin\left(\frac{1}{x}\right)dx$
bessel functionsclosed-formdefinite integralsimproper-integralsintegration
Related Solutions
We can generalize the integral by manipulating the Laplace transform of $J_{n}(bx)$, namely $$ \int_{0}^{\infty} J_{n}(bx) e^{-sx} \, dx = \frac{(\sqrt{s^{2}+b^{2}}-s)^{n}}{b^{n}\sqrt{s^{2}+b^{2}}}\ , \quad \ (n \in \mathbb{Z}_{\ge 0} \, , \text{Re}(s) >0 , \, b >0 )\tag{1}. $$
(See this question for a derivation of $(1)$ using contour integration.)
First let $s=p+ia$, where $p,a >0$.
A slight modification of the answer here shows that $\int_{0}^{\infty} J_{n}(bx) e^{-(p+ia)x} \, dx $ converges uniformly for all $p \in [0, \infty$).
This allows us to conclude that $$\begin{align} \int_{0}^{\infty} J_{n}(bx) e^{-iax} \, dx &= \lim_{p \downarrow 0}\int_{0}^{\infty} J_{n}(bx) e^{-(p+ia)x} \, dx \\ &= \lim_{p \downarrow 0} \frac{\left(\sqrt{(-p+ia)^2+b^{2}}-p-ia\right)^{n}}{b^{n}\sqrt{(p+ia)^2+b^{2}}} \\ &= \frac{\left(\sqrt{b^{2}-a^{2}}-ia\right)^{n}}{b^{n}\sqrt{b^{2}-a^{2}}}. \end{align}$$
So if $ a < b$, $$ \begin{align} \int_{0}^{\infty} J_{n}(bx) e^{-iax} \, dx &= \frac{\left(\sqrt{b^{2}-a^{2}+a^{2}} e^{-i \arcsin \left(\frac{a}{b}\right)}\right)^{n}}{b^{n} \sqrt{b^{2}-a^{2}}} \\ &= \frac{e^{-in \arcsin \left(\frac{a}{b}\right)}}{\sqrt{b^{2}-a^{2}}} .\end{align}$$
And if $a >b$, $$ \begin{align} \int_{0}^{\infty} J_{n}(bx) e^{-iax} \, dx &= \frac{\left(i\sqrt{a^{2}-b^{2}}-ia \right)^{n}}{b^{n}i \sqrt{a^{2}-b^{2}}} \\ &= \frac{-i e^{i \pi n /2} \left(\sqrt{a^{2}-b^{2}}-a \right)^{n}}{b^{n} \sqrt{a^{2}-b^{2}}}. \end{align}$$
Therefore,
$$\int_{0}^{\infty} J_{n}(bx) \sin(ax) \, dx = \begin{cases} \frac{\sin \left(n \arcsin \left(\frac{a}{b} \right) \right)}{\sqrt{b^{2}-a^{2}}} \, & \quad 0 < a < b \\ \frac{\cos \left(\frac{\pi n}{2} \right) \left(\sqrt{a^{2}-b^{2}} -a \right)^{n}}{b^{n} \sqrt{a^{2}-b^{2}}} & \quad a > b >0 \end{cases} $$
Here is a different way to set up that a parameter in order to apply Feynman's trick.$$I=2\int_0^\frac{\pi}{2} \ln\left(1+\sin^2t\right) dt\overset{t=\operatorname{arccot} x}=2\int_0^\infty \frac{\ln(2+x^2)-\ln(1+x^2)}{1+x^2}dx$$ Now let us consider $$I(a)=\int_0^\infty \frac{\ln(a+x^2)-\ln(1+x^2)}{1+x^2}dx\Rightarrow I'(a)=\int_0^\infty \frac{1}{(1+x^2)(a+x^2)}dx$$ $$=\frac{1}{a-1}\left(\int_0^\infty \frac{1}{x^2+1}dx-\int_0^\infty \frac{1}{x^2+a}dx\right)=\frac{1}{a-1}\left(\frac{\pi}{2}-\frac{1}{\sqrt a}\cdot \frac{\pi}{2}\right)=\frac{\pi}{2}\frac{1}{\sqrt a(1+\sqrt a)}$$ We are looking to find $I=2I(2)$, but since $I(1)=0$ we have: $$I=2\left(I(2)-I(1)\right)=2\int_1^2 I'(a)da=\pi \int_1^2 \frac{1}{\sqrt a(1+\sqrt a)}da$$ $$\overset {\large a=x^2}=2\pi\int_1^\sqrt 2 \frac{1}{1+x}dx=2\pi \ln(1+x)\bigg|_1^\sqrt 2=2\pi \ln\frac{1+\sqrt 2}{2}$$
Best Answer
I'm pretty sure I remember doing this one before...
Ah, there it is. AoPS link.
Quoting myself:
Doing it a second time? It took longer to find the old post than to reedit it for this site's formatting.