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.


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 $\sin$.

  • $\begingroup$ I don't see the point of your answer, just to say that $\frac1{x^{-2}+1}$ is analytic for $|x|<1$ ? $\endgroup$
    – reuns
    Jan 5 '20 at 1:54
  • $\begingroup$ @reuns To facilitate the OP’s understanding of your answer. $\endgroup$
    – Szeto
    Jan 5 '20 at 2:30

$u(t)$ is defined for $t>0$ but $$\int_0^\infty u(t)\cos(xt)dt=e^{-|x|}$$ is defined for all $x\in \Bbb{R}$. The inverse Fourier transform gives the unique solution $$u(t)=A(\frac1{t+i}+\frac1{-t+i}), t > 0$$ Expand in power series at $t=0$ you'll find that for $c\in (0,1)$, $t\in (0,1)$ it fits with $$\sum_{\Re(s) < 0} Res(t^{-s}\text{cosec}\bigg(\frac{s\pi}{2}\bigg),s)=\frac{1}{2\pi i}\int_{c-i\infty}^{c+i\infty} t^{-s}\text{cosec}\bigg(\frac{s\pi}{2}\bigg)ds$$ Then you need the analytic continuation theorem :

$\text{cosec}(\frac{s\pi}{2})$ has exponential decay on the vertical line $\Re(s)=c$ thus its inverse Fourier/Laplace/Mellin transform is analytic.

Whence its inverse Mellin transform is the analytic continuation of $\sum_{\Re(s) < 0} Res(t^{-s}\text{cosec}\bigg(\frac{s\pi}{2}\bigg),s)$ ie. it is $A(\frac1{t+i}+\frac1{-t+i})$

The whole thing is called Ramanujan master theorem.

  • $\begingroup$ Please explain further what you have written in the quotation. I am not familiar with this concept. I just know what is analytic continuation. $\endgroup$ Jan 4 '20 at 15:41
  • $\begingroup$ The inverse Fourier transform is given by an integral, the exponential decay implies this integral converges and is holomorphic and analytic not only for $t$ real but also for every $t\in \Bbb{C}$ with $|\Im(t)|$ not too large $\endgroup$
    – reuns
    Jan 4 '20 at 16:08
  • $\begingroup$ Ok. But why the inverse Mellin transform is the analytic continuation? $\endgroup$ Jan 4 '20 at 16:28
  • $\begingroup$ If you don't see that inverse Mellin transform and inverse Fourier transform are the same thing then you have a problem $\endgroup$
    – reuns
    Jan 4 '20 at 16:33

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