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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.

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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 $\sin$.

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  • $\begingroup$ 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
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$u(t)$ 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.

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  • $\begingroup$ 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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