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

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What have you tried? Looks like a classic case for expanding the sine using Euler's formula. – Henning Makholm Nov 7 '11 at 18:04
@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. – Arturo Magidin Nov 7 '11 at 18:09
@Henning Makholm: You're right. There is nothing to do after using Euler's formula. Thank you! – Gardel Nov 7 '11 at 18:11
@ Arturo Magidin: Thanks, I'm starting to use the forum and therefore need that kind of information. – Gardel Nov 7 '11 at 18:15
up vote 9 down vote accepted

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}

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

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

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

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.

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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 – Ross Millikan Nov 7 '11 at 18:56
thank you for finding the typo. – Valerio Capraro Nov 7 '11 at 19:24

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.

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