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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1$\begingroup$ What have you tried? Looks like a classic case for expanding the sine using Euler's formula. $\endgroup$– hmakholm left over MonicaNov 7, 2011 at 18:04
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$\begingroup$ @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 MagidinNov 7, 2011 at 18:09
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$\begingroup$ @Henning Makholm: You're right. There is nothing to do after using Euler's formula. Thank you! $\endgroup$– GardelNov 7, 2011 at 18:11
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$\begingroup$ @ Arturo Magidin: Thanks, I'm starting to use the forum and therefore need that kind of information. $\endgroup$– GardelNov 7, 2011 at 18:15
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4 Answers
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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. $\endgroup$– TapuNov 7, 2011 at 18:53
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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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1$\begingroup$ @anon: So, $\sint$ = imaginary part of $e^{it}$ simplifies things better. $\endgroup$– TapuNov 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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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 $\endgroup$ Nov 7, 2011 at 18:56
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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.
answered Nov 7, 2011 at 18:03