Differential equation with improper integral

Question

I have to find $f$ from the following equation: $$\int_0^\infty f(u)\sin(ut)\,du=e^{-t}$$ where $t>0$

Any hint? Can I use the Laplace Transform method?

Answer 1

If you are not interested in finding all the solutions but just a solution you may notice that the Laplace and Cauchy distributions are related in such a way that, for any $t>0$, $$ \int_{0}^{+\infty}\frac{u \sin(u t)}{1+u^2}\,du = \frac{1}{2}\int_{\mathbb{R}}\frac{u\sin(u t)}{1+u^2}\,du=\pi\,\text{Re}\operatorname*{Res}_{u=i}\frac{u e^{iut}}{1+u^2}=\frac{\pi}{2}e^{-t} $$ so $f(u)=\frac{2}{\pi}\cdot\frac{u}{1+u^2}$ is a solution.

Answer 2

With CAS help using MellinTransform:

$$\mathcal{M}_t\left[\int_0^{\infty } f(u) \sin (u t) \, du=\exp (-t)\right](s)$$ $$\Gamma (s) \sin \left(\frac{\pi s}{2}\right) \mathcal{M}_u[f(u)](1-s)=\Gamma (s)$$ $$\mathcal{M}_u[f(u)](1-s)=\frac{1}{\sin \left(\frac{\pi s}{2}\right)}$$ $$\mathcal{M}_u[f(u)](s)=\csc \left(\frac{1}{2} \pi (1-s)\right)$$ then inverse MellinTransform is: $$\mathcal{M}_s^{-1}\left[\mathcal{M}_u[f(u)](s)\right](u)=\mathcal{M}_s^{-1}\left[\csc \left(\frac{1}{2} \pi (1-s)\right)\right](u)=\mathcal{M}_s^{-1}\left[\sec \left(\frac{\pi s}{2}\right)\right](u)=\frac{2 u}{\pi \left(1+u^2\right)}$$

$$f(u)=\frac{2 u}{\pi \left(1+u^2\right)}$$