Prove that $\int_0^\infty \frac{\sin nx}{x}dx=\frac{\pi}{2}$

Question

There was a question on multiple integrals which our professor gave us on our assignment.

QUESTION: Changing order of integration, show that $$\int_0^\infty \int_0^\infty e^{-xy}\sin nx \,dx \,dy=\int_0^\infty \frac{\sin nx}{x}dx$$ and hence prove that $$\int_0^\infty \frac{\sin nx}{x}dx=\frac{\pi}{2}$$

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MY ATTEMPT: I was successful in proving the first part.

Firstly, I can state that the function $e^{-xy}\sin nx$ is continuous over the region $\mathbf{R}=\{(x,y): 0<x<\infty,0<y<\infty\}$

$$\int_0^\infty \int_0^\infty e^{-xy}\sin nx \,dx \,dy$$ $$=\int_0^\infty \sin nx \left\{\int_0^\infty e^{-xy}\,dy\right\} \,dx$$ $$=\int_0^\infty \sin nx \left[\frac{e^{-xy}}{-x}\right]_0^\infty \,dx$$ $$ =\int_0^\infty \frac{\sin nx}{x}dx$$

However, the second part of the question yielded a different answer.

$$\int_0^\infty \int_0^\infty e^{-xy}\sin nx \,dx \,dy$$ $$=\int_0^\infty \left\{\int_0^\infty e^{-xy} \sin nx \,dx\right\} \,dy$$ $$=\int_0^\infty \frac{ndy}{\sqrt{n^2+y^2}}$$

which gives an indeterminate result, not the desired one.

Where did I go wrong? Can anyone help?

Answer 1

You should have obtained $$\int_{x=0}^\infty e^{-yx} \sin nx \, dx = \frac{n}{n^2 + y^2}.$$ There are a number of ways to show this, such as integration by parts. If you would like a full computation, it can be provided upon request.

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Let $$I = \int e^{-xy} \sin nx \, dx.$$ Then with the choice $$u = \sin nx, \quad du = n \cos nx \, dx, \\ dv = e^{-xy} \, dx, \quad v = -\frac{1}{y} e^{-xy},$$ we obtain $$I = -\frac{1}{y} e^{-xy} \sin nx + \frac{n}{y} \int e^{-xy} \cos nx \, dx.$$ Repeating the process a second time with the choice $$u = \cos nx \, \quad du = -n \sin nx \, dx, \\ dv = e^{-xy} \, dx, \quad v = -\frac{1}{y} e^{-xy},$$ we find $$I = -\frac{1}{y}e^{-xy} \sin nx - \frac{n}{y^2} e^{-xy} \cos nx - \frac{n^2}{y^2} \int e^{-xy} \sin nx \, dx.$$ Consequently $$\left(1 + \frac{n^2}{y^2}\right) I = -\frac{e^{-xy}}{y^2} \left(y \sin nx + n \cos nx\right),$$ hence $$I = -\frac{e^{-xy}}{n^2 + y^2} (y \sin nx + n \cos nx) + C.$$ Evaluating the definite integral, for $y, n > 0$, we observe $$\lim_{x \to \infty} I(x) = 0, \quad I(0) = -\frac{n}{n^2 + y^2},$$ and the result follows.

Answer 2

You just need to integrate $$ \int_{0}^{\infty}e^{-xy}\sin[nx]dx=\frac{1}{2i}\int^{\infty}_{0}\left(e^{(ni-y)x}-e^{-(ni+y)x}\right)dx $$ And you use the fact $$ \int^{\infty}_{0}e^{cx}dx=\frac{1}{c}\Big|^{\infty}_{0}e^{cx}=\frac{-1}{c} $$ Thus you have $$ \frac{1}{2i}\left(\frac{1}{y-ni}-\frac{1}{y+ni}\right)=\frac{n}{n^2+y^2} $$ I am sure there are other ways to do this (intergration by parts, induction, differentiation under integral sign, etc). If I am not mistaken, this is one of many ways to prove the identity $\int^{\infty}_{0}\frac{sin[x]}{x}dx=\frac{\pi}{2}$ using dominated convergence theorem.

Answer 3

Since for $y>0$, $ e^{x(-y+in)}\to 0~~as~~x\to \infty$ we have,

$$\int_0^\infty e^{-xy}\sin nx \,dx =Im\left(\int_0^\infty e^{x(-y+in)} \,dx\right) = Im\left(\frac{1}{y-in} \right) = \frac{n}{y^2 +n^2} $$

Thus$$\int_0^\infty \int_0^\infty e^{-xy}\sin nx \,dxdy =\int_0^\infty \frac{n}{y^2 +n^2}dy \overset{y=nu}{=}\int_0^\infty \frac{1}{u^2 +1}du =\frac{\pi}{2}$$