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Consider the double-integral $$ F(x,y) := \int_0^\infty d a \int_0^\infty d b\ \frac{ \sin(x a) \sin( y b ) }{a+b} \ . $$ By playing around in Mathematica I have come to believe that the above evaluates to $$ F(x,y) \stackrel{?}{=} \frac{\pi}{2(x+y)} \ . $$ How would one prove this result?

EDIT: I've been able to evaluate the $a$-integral, showing that the above is equal to $$ F(x,y) = \int_0^\infty db\ \sin(by) \bigg[ \text{Ci}(bx) \sin(bx) + \frac{1}{2} \cos(bx) \big( \pi - \text{Si}(bx) \big) \bigg] \ , $$ where $\text{Ci}$ and $\text{Si}$ are the cosine integral and sine integral functions, respectively. From here I cannot make progress.

There is a symmetry in the order of integration (ie. the above would look the same if I did the $b$-integral first). It is curious that the function seems to depend on $x+y$: I don't see how this dependence falls out of the definition of $F$.

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  • $\begingroup$ @TheSimpliFire See my edit above. $\endgroup$ – QuantumEyedea Jan 20 at 15:22
  • $\begingroup$ I think it follows the trick considering $\frac{1}{a+b}=\int_{0}^{+\infty}exp(-(a+b)c)dc$, using Fubini's theorem. $\endgroup$ – Oolong milk tea Jan 20 at 15:39
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$$I=\int_{0}^{\infty} \int_{0}^{\infty} \frac{\sin ax \sin bx}{a+b} da db = \int_{0}^{\infty} \int_{0}^{\infty} \sin ax \sin by ~ da ~ db ~ e^{-(a+b)t}~dt$$ $$I=\int_{0}^{\infty} dt\left(\int_{0}^{\infty} \sin ax ~e^{-ta} da \int_{0}^{\infty}\sin by ~ db ~ e^{-tb}\right)$$ $$I=\int_{0}^{\infty} \frac{xy}{(t^2+x^2)(t^2+y^2)}dt= \frac{xy}{y^2-x^2} \int_{0}^{\infty}\left(\frac{1}{t^2+x^2}-\frac{1}{t^2+y^2}\right)=\frac{\pi}{2(x+y)}.$$ Note that: $$J=\int \sin ax ~e^{-ta} da=\Im \int_{0}^{\infty} e^{-(t-ix)}=\frac{x}{t^2+x^2}$$

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    $\begingroup$ +1. maybe a sentence or two justifying the order of integrations? $\endgroup$ – dezdichado Jan 20 at 15:46
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$$F(x,y)=\int_0^\infty\int_0^\infty\frac{\sin(xa)\sin(yb)}{a+b}da\,db$$ $u=xa,da=du/x$ and $v=yb,db=dv/y$ so: $$F(x,y)=\int_0^\infty\int_0^\infty\frac{\sin u\sin v}{u/x+v/y}\frac{du}{x}\frac{dv}{y}=\int_0^\infty\int_0^\infty\frac{\sin u\sin v}{yu+xv}du\,dv$$ makes the sines simpler but we are probably going to have to use another system I think


$$F(x,y)=\int_0^\infty\int_0^\infty\frac{\sin(xa)\sin(yb)}{a+b}da\,db$$ we could try $u=a+b, v=a-b$ we just have to make sure the jabobian is non-zero: $$da\,db=\left|\frac{\partial(a,b)}{\partial(u,v)}\right|du\,dv$$ we can work out that: $$a=\frac{u+v}{2}$$ $$b=\frac{u-v}{2}$$ and so: $$\left|\frac{\partial(a,b)}{\partial(u,v)}\right|=\left|\begin{matrix}\frac12 & \frac12 \\ \frac12 &-\frac12\end{matrix}\right|=-\frac12$$ $$da\,db=-\frac12du\,dv$$ so our integral will become: $$F(x,y)=-\frac12\iint\limits_S\frac{\sin\left(\frac x2(u+v)\right)\sin\left(\frac y2(u-v)\right)}{u}dudv$$ Now you need to separate $u,v$ using compound angle rule and work out the domain $S$

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