# How do I solve the integral equation $\int_0^\infty f(y)\sin(xy)dy=e^{-x} , x>0$?

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?

• It may help from Euler's formula that $$\sin a = \frac{e^{ia} - e^{-ia}}{2i}$$ Jan 3 '20 at 17:03

Note that as $$x \to 0+$$, $$\sin(xy) \to 0$$ so if $$f$$ is integrable, you should have $$\int_0^\infty f(y) \sin(xy)\; dy \to 0$$ by Dominated Convergence. But of course $$e^{-x}$$ goes to $$1$$, not $$0$$. So there is certainly no integrable solution.

I suppose it's possible that there is a non-integrable solution, but that makes convergence of your integral problematic. Do you have any reason to expect a solution to exist?

• Yes, this was a question on a last year exam! I'll check if I saw it right Jan 4 '20 at 2:10
• Yes I saw it right...maybe my professor got something wrong? The question says explicitly to calculate f by solving the integral equation.... Jan 4 '20 at 2:12
• The question also says to prove that f belongs to L2. I don't know if that helps. Jan 4 '20 at 2:15
• Maybe the exam paper is wrong, if the sine changes to sin(x-y) does it have a solution? I'll try this tomorrow Jan 4 '20 at 2:31

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.

• Under the assumption that $x\geq 0$ that is =) Jan 4 '20 at 13:21

If $$f_o$$ is the odd extension of $$f$$ to $$\mathbb{R}$$, then $$e^{-x} = \int_0^{\infty}f(y)\sin(xy)dy \\ = \frac{1}{2}\int_{-\infty}^{\infty}f_o(y)\sin(xy)dy \\ = \frac{1}{2i}\int_{-\infty}^{\infty}f_o(y)e^{ixy}dy,$$ which works out by residues if you close in the upper half plane and you use $$f_o(y)=\frac{1}{\pi}\left(\frac{1}{y-i}+\frac{1}{y+i}\right) = \frac{2}{\pi}\frac{y}{y^2+1}.$$ As anticipated by Robert Israel, $$f(y)$$ is not integrable on $$(0,\infty)$$. The solution is valid when interpreted as an improper integral at $$\infty$$, which fits nicely with Complex Analysis and with Fourier Analysis. Any real function that is locally integrable on $$[0,\infty)$$ and is eventually monotonically increasing or decreasing to $$0$$ has a Fourier transform because the sine and cosine transforms can be interpreted as sums of integrals over a finite interval plus an alternating series with a general term that tends to $$0$$.