I'm trying to solve this integral equation, but I can't deal with the sine!
$$\int_0^\infty f(y)\sin(xy)dy=e^{-x} , x>0$$
How do I proceed?
I tried using the fourier transform but then I can't deal with the fourier transform of the sine.
The definition of Fourier transform I'm using is:
$$f(\xi)=\frac{1}{\sqrt{2\pi}}\int_{-\infty}^\infty f(x)e^{-i\xi x}dx$$
EDIT: I think I know what my problem was, I was doing the fourier transform of the sine individualy instead of doing the fourier transform of $f(y)\times sin(xy)$. Doing this and substituing the sine for its exponential formula will give the answer right?
Best Answer
The solution is $$f(y)=\frac{2}{\pi}\frac{y}{1+y^2}$$ I will add the derivation soon.
Derivation:
Let $$g(y)= \begin{cases} f(y)&&y\ge0 \\ 0&&y<0 \end{cases} $$ Then, with Euler's formula, we can rephrase the problem as $$\int^\infty_{-\infty}g(y)\left(\frac{e^{ixy}-e^{-ixy}}{2i}\right)dy=e^{-x}$$
Equivalently, $$\mathcal F\{g(y)\}(-x)-\mathcal F\{g(y)\}(x)=i\sqrt{\frac2\pi}e^{-x}$$ $$G(-x)=G(x)+i\sqrt{\frac2\pi}e^{-x}\qquad{x>0}$$
This functional equation does not tell much, as the solution is not unique. The best we can do is defining $$G(x)= \begin{cases} \varphi (x) && x>0 \\ \varphi(-x)+i\sqrt{\frac2\pi}e^{x} && x<0 \end{cases} $$ for some $\varphi (x):\mathbb R^+$ with sufficiently nice properties.
Then, $$ \begin{align} g(y)&=\frac{1}{\sqrt{2\pi}}\int^\infty_{-\infty}G(x)e^{ixy}dx \\ \sqrt{2\pi}g(y)&=\int^\infty_{0}G(x)e^{ixy}dx+\int^{\infty}_{0}G(-x)e^{-ixy}dx \\ &=\int^\infty_{0}\varphi(x)e^{ixy}dx+\int^{\infty}_{0}\left(\varphi(x)+i\sqrt{\frac2\pi}e^{-x}\right)e^{-ixy}dx \\ &=2\int^\infty_{0}\varphi(x)\cos(xy)dx+i\sqrt{\frac2\pi}\int^{\infty}_{0}e^{-x}e^{-ixy}dx \\ &=2\int^\infty_{0}\varphi(x)\cos(xy)dx+i\sqrt{\frac2\pi}\frac1{1+iy} \\ g(x)&=\sqrt{\frac2\pi}\int^\infty_{0}\varphi(x)\cos(xy)dx+\frac i\pi\frac1{1+iy} \\ \end{align} $$
Let $a>0$. By assumption, $g(-a)=0$. Therefore, $$\sqrt{\frac2\pi}\int^\infty_{0}\varphi(x)\cos(ay)dx+\frac i\pi\frac1{1-ia}=0$$
Then, $$\begin{align} g(a)&=\sqrt{\frac2\pi}\int^\infty_{0}\varphi(x)\cos(ay)dx+\frac i\pi\frac1{1+ia} \\ &=-\frac i\pi\frac1{1-ia}+\frac i\pi\frac1{1+ia} \\ &=\frac2\pi\frac{a}{1+a^2} \end{align} $$
Hence, $$f(y)=\frac{2}{\pi}\frac{y}{1+y^2}$$
Verification:
(The calculations below are a bit sloppy as I assumed that the integral and differentiation can be interchanged.)
$$\begin{align} L.H.S. &=\frac2\pi \int^\infty_0\frac{y}{1+y^2}\sin(xy)dy \\ &=-\frac2\pi\frac{\partial}{\partial x}\int^\infty_0\frac{\cos(xy)}{1+y^2}dy \\ &=-\frac1\pi\frac{\partial}{\partial x}\int^\infty_{-\infty}\frac{\cos(xy)}{1+y^2}dy \\ &=-\frac1\pi\frac{\partial}{\partial x}\pi e^{-x} \\ &=e^{-x}\\ &=R.H.S. \end{align} $$
Here we utilized the well-known integral identity $$\int^\infty_{-\infty}\frac{\cos(ax)}{1+x^2}dx=\pi e^{-|a|}$$ For its proof, see the accepted answer here.