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

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?

SchrodingersCat
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    For one thing, you needlessly drag along the parameter $n$. It is silently assumed that $n>0$ -- since if $n=0$ or $n<0$, the integral $\int_0^\infty (sin(nx)/x),dx$ is $0$ resp. $-\pi/2$. For $n>0$ you have $\int_0^\infty (sin(nx)/x),dx =\int_0^\infty (sin(nx)/(nx)),d(nx) = \int_0^\infty (sin(x)/x),dx$. – chizhek May 22 '16 at 11:17

4 Answers4

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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.

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.

heropup
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  • Thanks, I knew the result. But I don't know how I added the square root. Thanks... – SchrodingersCat May 22 '16 at 07:39
  • There is an additional 'minus sign' in the $I(0)$ you just proved. Can you justify it please? – SchrodingersCat May 22 '16 at 07:49
  • @SchrodingersCat The sign of $I(0)$ is correct as written. Note that the definite integral has value $$I(\infty) - I(0) = 0 - \left( - \frac{n}{n^2 + y^2} \right) = \frac{n}{n^2 + y^2},$$ as desired. – heropup May 22 '16 at 08:29
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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.

Bombyx mori
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Another approach to evaluating \begin{equation} \int\limits_{0}^{\infty} \mathrm{e}^{-yx} \sin(nx) \mathrm{d} x \end{equation} is to recognize that this expression is the Laplace transform of $f(x) = \sin(nx)$

Thus, \begin{equation} \int\limits_{0}^{\infty} \mathrm{e}^{-yx} \sin(nx) \mathrm{d} x = \mathcal{L}[\sin(nx)](y) = \frac{n}{n^2 + y^2} \end{equation}

Or

Let $$\sin(nx) = \frac{\mathrm{e}^{inx}-\mathrm{e}^{-inx}}{i2}$$ \begin{equation} \int\limits_{0}^{\infty} \mathrm{e}^{-yx} \sin(nx) \mathrm{d} x = \frac{1}{i2} \int\limits_{0}^{\infty} \left[\mathrm{e}^{(-y + in)x} - \mathrm{e}^{(-y - in)x} \right] \mathrm{d} x \\ = \frac{1}{i2} \left( \frac{\mathrm{e}^{(-y + in)x}}{-y + in} - \frac{\mathrm{e}^{(-y - in)x}}{-y - in} \right) |_{0}^{\infty} \\ = \frac{1}{i2} \left( \frac{1}{-y - in} - \frac{1}{-y + in} \right) \\ = \frac{n}{n^2 + y^2} \end{equation}

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

Guy Fsone
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