8
$\begingroup$

Without using Laplace transforms, how do I show that for every positive number $x$ the following equation is valid? $$\int_{0}^{\infty}e^{-xt}\sin(t)dt=\frac{1}{x^2+1}. $$

$\endgroup$
  • 1
    $\begingroup$ What have you tried? Looks like a classic case for expanding the sine using Euler's formula. $\endgroup$ – Henning Makholm Nov 7 '11 at 18:04
  • $\begingroup$ @Gardel: Generally speaking, it is a good idea to avoid (i) displayed equations in the title; and (ii) titles that consist entirely of $\LaTeX$. This has to do with the way they are displayed and handled by the site, as well as issues with searches. $\endgroup$ – Arturo Magidin Nov 7 '11 at 18:09
  • $\begingroup$ @Henning Makholm: You're right. There is nothing to do after using Euler's formula. Thank you! $\endgroup$ – Gardel Nov 7 '11 at 18:11
  • $\begingroup$ @ Arturo Magidin: Thanks, I'm starting to use the forum and therefore need that kind of information. $\endgroup$ – Gardel Nov 7 '11 at 18:15
9
$\begingroup$

Yet another one (particularly useful when evaluating particular integral of ODE):

\begin{align} \int_{0}^{\infty}e^{-xt}\sin t\ dt&=\Im\left[\int_0^\infty e^{-xt}e^{it}\ dt\right]\\ &=\Im\left[\int_0^\infty e^{-(x-i)t}\ dt\right]\\ &=\Im\left[\frac1{x-i}\right]\\ &=\Im\left[\frac1{x-i}\cdot\frac{x+i}{x+i}\right]\\ &=\frac1{x^2+1}. \end{align}

$\endgroup$
  • $\begingroup$ Please note that same approach can be applied to get integral involving $\sin mt$, $\cos mt$ etc. $\endgroup$ – Tapu Nov 7 '11 at 18:53
  • $\begingroup$ This is really amazing, thank you... $\endgroup$ – Gardel Nov 7 '11 at 21:14
7
$\begingroup$

$\rm\bf Hint$: $$\large e^{it}=\cos(t)+i\sin(t)\implies \sin(t)=\frac{e^{it}-e^{-it}}{2i}.$$

$\endgroup$
  • $\begingroup$ yes, now I see. Thank you! $\endgroup$ – Gardel Nov 7 '11 at 18:07
  • 1
    $\begingroup$ @anon: So, $\sint$ = imaginary part of $e^{it}$ simplifies things better. $\endgroup$ – Tapu Nov 7 '11 at 19:02
6
$\begingroup$

Why the "complex analysis" tag? It's just a double integration:

$$ \int e^{-xt}\sin(t)dt=-\frac{1}{x}e^{-xt}\sin(t)+\frac{1}{x}\int e^{-xt}\cos(t)dt= $$

$$ =-\frac{1}{x}e^{-xt}\sin(t)-\frac{1}{x^2}e^{-xt}\cos(t)-\frac{1}{x^2}\int e^{-xt}\sin(t)dt $$

It follows that

$$ \left(1+\frac{1}{x^2}\right)\int e^{-xt}\sin(t)dt= -\frac{1}{x}e^{-xt}\sin(t)-\frac{1}{x^2}e^{-xt}\cos(t)+ $$

It follows that

$$ \left(1+\frac{1}{x^2}\right)\int_0^\infty e^{-xt}\sin(t)dt= \frac{1}{x^2} $$

$$ \left(\frac{x^2+1}{x^2}\right)\int_0^\infty e^{-xt}\sin(t)dt= \frac{1}{x^2} $$

$$ \int_0^\infty e^{-xt}\sin(t)dt= \frac{1}{1+x^2} $$ It follows that the integral is equal to $\displaystyle \frac{1}{x^2+1}$, as you wrote.

$\endgroup$
  • 3
    $\begingroup$ When you find d(cos(t)) in the second line it has a minus sign, so the last term has the wrong sign. Please put the $dt$'s in as well $\endgroup$ – Ross Millikan Nov 7 '11 at 18:56
  • $\begingroup$ thank you for finding the typo. $\endgroup$ – Valerio Capraro Nov 7 '11 at 19:24
4
$\begingroup$

If you integrate by parts twice, using $u=\exp(-xt), dv=\sin(t)dt$ then $dv= \cos(t)dt$ you get the same integral back and can solve for it.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.