Solve the following singular integral equation using suitable integral
transform : $$\int_0^\infty u(t)\cos(xt)dt=e^{-x}$$
One easy method is if I use fourier cosine transform. But instead I chose to apply Mellin transform, to see what happens next. We know that $$\mathcal{M}\bigg(\int_0^\infty u(t)\cos(xt)dt;s\bigg)=U(1-s)\Gamma(s)\cos\bigg(\frac{s\pi}{2}\bigg)$$ where $U(s)=\mathcal{M}(u(x);s)$ and $$\mathcal{M}(e^{-x};s)=\Gamma(s)$$ Now, we have fom the given equation $$U(1-s)\Gamma(s)\cos\bigg(\frac{s\pi}{2}\bigg)=\Gamma(s)$$ $$\implies U(s)=\text{cosec}\bigg(\frac{s\pi}{2}\bigg)$$ Taking Mellin inverse $$u(x)=\frac{1}{2\pi i}\int_{c-i\infty}^{c+i\infty} x^{-s}\text{cosec}\bigg(\frac{s\pi}{2}\bigg)ds$$
Now I can't apply the method of residues as the cosecant term being present in the integrand. How to evaluate the above complex integral? Any help is appreciated.
Best Answer
Clearly, the Mellin transform identity for $\cos$ is valid only when $0<\Re s<1$. Hence, we are forced to choose $c\in (0,1)$ for Mellin inverse.
Let’s first solve for $u(x)$ when $0<x<1$.
In this case, the integrand decays exponentially on the left half plane due to $x^s=e^{(\ln x)s}$. Hence, we choose a contour from $c-i\infty$ to $c+i\infty$, and close the contour by attaching a infinitely big semicircle on its left.
Obviously, the integral over the arc vanishes. Therefore, by residue theorem, the Mellin inverse equals $$\sum\text{residues of $x^{-s}\csc\frac{\pi s}2$ on the left half plane}$$
Note that singularities enclosed are simple poles and are at $s=-2n$, $n=0,1,2,\cdots$. The residue is $$x^{2n}\lim_{s\to -2n}(s+2n)\csc\frac{\pi s}2=\frac{2}{\pi}(-1)^n x^{2n}$$
Summing the residue from $n=0$ to $\infty$, we found that the Mellin inverse is $$u(x)=\frac{2}{\pi}\frac{1}{x^2+1}$$
It is (un)surprising that we get the same result for $x>1$ (semicircle on the right half plane) - this is an instance of analytic continuation. Hence, we conclude that $\frac{2}{\pi}\frac{1}{x^2+1} $ is the solution of $u(x)$ for all $x>0$.
Note that this question is highly similar to another one I recently answered. The two different solutions are respectively the real and imaginary parts of $\frac 2\pi \frac1{1-ix}$, depending on whether it is a $\cos$ or a $\sin$.