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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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    $\begingroup$ What have you tried? Looks like a classic case for expanding the sine using Euler's formula. $\endgroup$ Nov 7, 2011 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. $\endgroup$ Nov 7, 2011 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, 2011 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, 2011 at 18:15

4 Answers 4

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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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  • $\begingroup$ Please note that same approach can be applied to get integral involving $\sin mt$, $\cos mt$ etc. $\endgroup$
    – Tapu
    Nov 7, 2011 at 18:53
  • $\begingroup$ This is really amazing, thank you... $\endgroup$
    – Gardel
    Nov 7, 2011 at 21:14
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$\rm\bf Hint$: $$\large e^{it}=\cos(t)+i\sin(t)\implies \sin(t)=\frac{e^{it}-e^{-it}}{2i}.$$

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  • $\begingroup$ yes, now I see. Thank you! $\endgroup$
    – Gardel
    Nov 7, 2011 at 18:07
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    $\begingroup$ @anon: So, $\sint$ = imaginary part of $e^{it}$ simplifies things better. $\endgroup$
    – Tapu
    Nov 7, 2011 at 19:02
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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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    $\begingroup$ 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 $\endgroup$ Nov 7, 2011 at 18:56
  • $\begingroup$ thank you for finding the typo. $\endgroup$ Nov 7, 2011 at 19:24
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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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