Xrfgtjtk

I have stumbled onto an interesting integral$$int_0^infty sin(x)sinleft(frac{1}{x}right)dx$$ 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.

I'm pretty sure I remember doing this one before...

Ah, there it is. AoPS link.

Quoting myself:

Well, you can't just leave it there. This one definitely deserves more than just an answer.

Now, trying to work out an actual solution - $1$ is not a special point for $sin$. There's nothing special about picking $frac1x$ there - so, to give ourselves more handles to work with, we stretch the problem. Define
$$F(a) = int_0^{infty}sin(x)sinleft(frac{a}{x}right),dx$$
Now, we would like to differentiate $F$. Differentiating under the integral sign gets us $F'(a) stackrel{?}{=} int_0^{infty}frac1xsin(x)cosleft(frac{a}{x}right),dx$ and then $F''(a) stackrel{?}{=} int_0^{infty}frac{-1}{x^2}sin(x)sinleft(frac{a}{x}right),dx$, while substituting $t=frac{a}x$ gets us $F(a) = int_{infty}^{0}sinleft(frac{a}{t}right)sin(t)cdotfrac{-a}{t^2},dt stackrel{?}{=} -aF''(a)$.

Why the question marks? Because I'm doing things with conditionally convergent integrals that I would need absolute convergence to justify.

It's still informative. That's not the Bessel differential equation, but it's related - looking at the power series, we would get a Bessel function of $2sqrt{a}$. In light of that, let's redefine:
$$G(a) = int_0^{infty}sin(x)sinleft(frac{a^2}{x}right),dx$$
Now, we need to massage this into a form that will let us justify the differentiation. First, split at $x=a$ and fold with that substitution:
begin{align*}G(a) &= int_0^a sin(x)sinleft(frac{a^2}{x}right),dx + int_a^{infty} sin(x)sinleft(frac{a^2}{x}right),dx\
&= int_0^a sin(x)sinleft(frac{a^2}{x}right),dx + int_a^0 sinleft(frac{a^2}{t}right)sin(t)cdotfrac{-a^2}{t^2},dt\
&= int_0^a sin(x)sinleft(frac{a^2}{x}right),dx + int_0^a frac{a^2}{x^2}sin(x)sinleft(frac{a^2}{x}right),dx\
G(a) &= int_0^a left(1+frac{a^2}{x^2}right)sin(x)sinleft(frac{a^2}{x}right),dxend{align*}
This form is improper at zero due to the rapid growth (balanced by oscillation), so we integrate by parts:
begin{align*}G(a) &= int_0^a left(1+frac{a^2}{x^2}right)sin(x)sinleft(frac{a^2}{x}right),dx\
&,_{dv = tfrac{a^2}{x^2}sintfrac{a^2}{x},dx,quad v = costfrac{a^2}{x}}^{u = (tfrac{x^2}{a^2}+1)sin x,quad du = (tfrac{x^2}{a^2}+1)cos x + tfrac{2x}{a^2}sin x,dx}\
G(a) &= left[left(frac{x^2}{a^2}+1right)sin xcosfrac{a^2}{x}right]_{x=0}^{x=a} - int_0^a cosfrac{a^2}{x}left(left(frac{x^2}{a^2}+1right)cos x + frac{2x}{a^2}sin xright),dx\
&= sin(2a) - int_0^a cosfrac{a^2}{x}left(left(frac{x^2}{a^2}+1right)cos x + frac{2x}{a^2}sin xright),dxend{align*}
This form is now absolutely convergent; it's the integral of a bounded function on a bounded interval, continuous except at one endpoint. If we differentiate it - well, then we get terms in which we multiply by $frac{2a}{x}$, and it's improper again. Only the $cosfrac{a^2}{x}cos x$ term causes trouble, so let's separate that out:
$$G(a) = sin(2a) - int_0^a cosfrac{a^2}{x}left(frac{x^2}{a^2}cos x + frac{2x}{a^2}sin xright),dx - int_0^a cosfrac{a^2}{x}cos x,dx$$
begin{align*}int_0^a cosfrac{a^2}{x}cos x,dx &= left[-frac{x^2}{a^2}cos xsinfrac{a^2}{x}right]_{x=0}^{x=a} + int_0^a sinfrac{a^2}{x}left(frac{2x}{a^2}cos x - frac{x^2}{a^2}sin(x)right),dx\
G(a) &= frac32sin(2a) -int_0^a frac{x^2}{a^2}cosleft(x+frac{a^2}{x}right) + frac{2x}{a^2}sinleft(x+frac{a^2}{x}right),dxend{align*}
That's a lot of trig identities that go into that last line - mostly the angle-sum identities, which our manipulation was nice enough to set up for us cleanly.

And now, we can finally differentiate cleanly.
begin{align*}G(a) &= frac32sin(2a) -int_0^a frac{x^2}{a^2}cosleft(x+frac{a^2}{x}right) + frac{2x}{a^2}sinleft(x+frac{a^2}{x}right),dx\
G'(a) &= 3cos(2a) - cos(2a) - frac{2}{a}sin(2a)\
&quad+int_0^a frac{2x^2}{a^3}cosleft(*right) + frac{2x}{a}sinleft(*right) + frac{4x}{a^3}sinleft(*right) - frac{4}{a}cosleft(*right),dx\
G'(a) &= 2cos(2a) -frac{2}{a}sin(2a)\
&quad + int_0^a left(frac{2x^2}{a^3}-frac{4}{a}right)cosleft(x+frac{a^2}{x}right) + left(frac{2x}{a} + frac{4x}{a^3}right)sinleft(x+frac{a^2}{x}right),dxend{align*}
And now... we'd like to differentiate again, but one of the terms doesn't have any powers of $x$ to cushion it, and differentiating it will break absolute convergence. Separate it out and integrate by parts again:
begin{align*}int_0^a frac{x^2}{a^3}left(1-frac{a^2}{x^2}right)cosleft(x+frac{a^2}{x}right),dx &= left[frac{x^2}{a^3}sinleft(x+frac{a^2}{x}right)right]_{x=0}^{x=a} - int_0^a frac{2x}{a^3}sinleft(x+frac{a^2}{x}right),dx\
&= frac{1}{a}sin(2a) - int_0^a frac{2x}{a^3}sinleft(x+frac{a^2}{x}right),dxend{align*}
Multiply by $4$ and add/subtract, to clear the $frac{4}{a}cos(*)$ term:
begin{align*} G'(a) &= 2cos(2a) + frac{2}{a}sin(2a) + int_0^a -frac{2x^2}{a^3}cosleft(x+frac{a^2}{x}right) + left(frac{2x}{a} - frac{4x}{a^3}right)sinleft(x+frac{a^2}{x}right),dx\
G'(a) &= 2cos(2a) - frac1asin(2a) + frac2a G(a) + int_0^a frac{2x}{a}sinleft(x+frac{a^2}{x}right),dx\
G''(a) &= -4sin(2a) - frac2acos(2a) + frac1{a^2}sin(2a) -frac2{a^2} G(a) + frac2a G'(a) + 2sin(2a)\
&quad + int_0^a 4cos(*) - frac{2x}{a^2}sin(*),dx\
G''(a) &= -2sin(2a) + frac1{a^2}sin(2a) - frac2acos(2a) -frac2{a^2} G(a) + frac2a G'(a)\
&quad + int_0^a 4cos(*) - frac{2x}{a^2}sin(*),dxend{align*}
Now, we deal with those integrals. From an integration by parts earlier, $int_0^a (x^2-a^2)cos(*) + 2xsin(*),dx = a^2sin(2a)$. From our expression for $G$, $int_0^a x^2cos(*) + 2xsin(*) = frac{3a^2}{2}sin(2a)-a^2 G(a)$. Subtract the two, divide by $a^2$, and $int_0^a cos(*),dx = frac12sin(2a) - G(a)$. Similarly, from our latest expression for $G'$, $int_0^a xsin(*),dx = frac{a}{2}G'(a) - G(a) - acos(2a) + frac12sin(2a)$. Apply these to the formula for $G''$, and
begin{align*}G''(a) &= -2sin(2a) + frac1{a^2}sin(2a) - frac2acos(2a) - frac2{a^2}G(a) + frac2{a}G'(a) +\
&quad 2sin(2a) - 4 G(a) - frac1a G'(a) + frac2{a^2}G(a) + frac2acos(2a) - frac1{a^2}sin(2a)\
G''(a) &= frac1{a}G'(a) - 4 G(a) end{align*}
This almost looks like a stretched Bessel equation of order zero - but the middle term has the wrong sign, and that's not something we can change by just flipping the sign of $x$ or something.
No, the way to change that middle term is to multiply the function by powers of $x$. Combining with that stretch, let $H(t)=frac1t Gleft(frac{t}{2}right)$. Then $H'(t)=frac{1}{2t}G'left(frac{t}{2}right)-frac1{t^2}Gleft(frac{t}{2}right)$, $H''(t)=frac{1}{4t}G''left(frac{t}{2}right)-frac1{t^2}G'left(frac{t}{2}right)+frac{2}{t^3}Gleft(frac{t}{2}right)$, and
begin{align*}t^2 H''(t) + t H(t) + (t^2-1)H(t) &= frac{t}{4}G''left(frac{t}{2}right)-G'left(frac{t}{2}right)+frac2t Gleft(frac{t}{2}right) + frac12G'left(frac{t}{2}right)\
&quad - frac1t Gleft(frac{t}{2}right) + t Gleft(frac{t}{2}right) - frac1t Gleft(frac{t}{2}right)\
&= frac{t}{4}G''left(frac{t}{2}right) - frac12 G'left(frac{t}{2}right) + tGleft(frac{t}{2}right)\
&stackrel{s=t/2}{=} frac{s}{2}left(G''(s) -frac1s G'(s) + 4 G(s)right) = 0end{align*} That equation for $H(t)$ is the Bessel equation of order $1$.

Now, we still need an initial condition. What is $G(0)$? Well, our equation degenerates there in several ways, but we can take a limit:
begin{align*}lim_{ato 0^+}G(a) &= lim_{ato 0^+}frac32sin(2a) -lim_{ato 0^+}int_0^a frac{x^2}{a^2}cosleft(x+frac{a^2}{x}right) + frac{2x}{a^2}sinleft(x+frac{a^2}{x}right),dx\
&= 0 - lim_{ato 0^+}int_0^1 left(t^2cosleft(at+frac{a}{t}right) + frac{2t}{a}sinleft(at+frac{a}{t}right)right)cdot a,dt = 0end{align*}
The first part of that integral is bounded by $at^2$, which tends to zero uniformly. For the second part, we use $|sin(x)|le |x|$; it's bounded by $2at^2 + 2a$, which also goes to zero uniformly.
But that's not enough - we're going to divide by $x$ at least once. We go one step stronger, and look at $lim_{ato 0^+}frac1a G(a)$. The $sin$ term gets us $3$. In the integral term, the integrand tends to $t^2cos(0) + 2t^2+2$ pointwise; by dominated convergence (using the same estimates as before), $lim_{ato 0^+}frac1a G(a) = 3 - 3 = 0$. That... well, that tells us $H$ is a multiple of the standard Bessel function $J_1$. We need to go another degree higher to find what multiple.
For that, the expression we're working with isn't terribly convenient, so we go back to a much earlier one:
begin{align*}G(a) &= int_0^a left(1+frac{a^2}{x^2}right)sin(x)sinleft(frac{a^2}{x}right),dx\
frac1{a^2}G(a) &= int_0^1 left(frac1{a^2}+frac1{a^2t^2}right)sin(at)sinfrac{a}{t}cdot a,dt\
frac1{a^2}G(a) &approx int_0^1 frac{t^2+1}{at^2}cdot atcdot sinfrac{a}{t},dt\
frac1{a^2}G(a) &approx int_0^1 left(t+frac1tright)sinfrac{a}{t},dtend{align*}
As $ato 0^+$, that last form of the integrand tends to zero. The $tsinfrac{a}{t}$ term tends to zero uniformly, while the other term is large enough to matter at zero. Now, we substitute:
begin{align*}frac1{a^2}G(a) &approx int_0^1 frac1tsinfrac{a}{t},dt\
&,^{t=tfrac{a}{x}}_{dt=-tfrac{a}{x^2},dx}\
frac1{a^2}G(a) &approx int_{infty}^{a} -frac{x}{a}sin(x)cdotfrac{a}{x^2},dx\
frac1{a^2}G(a) &approx int_a^{infty} frac{sin x}{x},dx = frac{pi}{2}end{align*}
That's our limit - $lim_{ato 0^+} frac1{a^2}G(a)=frac{pi}{2}$. For normalization, note that $J_1(t)approx frac{t}{2}$ for $t$ near zero. Then $H(2a)approx frac1{2a}G(a) approx frac{pi}{4}a$, and thus $H(2a)=frac{pi}{4}J_1(2a)$. Converting back to $G$, $G(a) = 2a H(2a) = frac{pi a}{2}J_1(2a)$.
The original question asked was $G(1)$, for an integral of $frac{pi}{2} J_1(2)$. The answer is confirmed... and it only took me most of a day to get there.

Doing it a second time? It took longer to find the old post than to reedit it for this site's formatting.

Maple gets the answer
$$ frac{pi}{2} J_1(2) $$
where $J_1$ is the Bessel function of first order. It says it used the "meijerg" method.

Consider
$$F(a,b) = int_0^{infty}sin(ax)sinleft(frac{b}{x}right),dx$$

Now, the Laplace transform of $F$ as a function of $b$,

$$mathcal{L}(F)=int_0^{infty}frac{xsin(ax)}{1+s^2x^2}dx=frac{pi}{2}frac{e^{-frac{a}{s}}}{s^2}$$

The last integral is computed in this site many times.

Finally, to invert the Laplace transform to recover $F(a,b)$ we use the inspection method.

That means that we'll look at the table Laplace transform pairs.

In this case we have good luck,

$$F(a,b) =frac{pi}{2}sqrtfrac{b}{a}J_1(2sqrt{ab})$$

where $J_1(x)$ is the Bessel Function of the First Kind