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}. $
How can we evaluate $\int_{0}^{\infty}e^{-xt}\sin(t)dt$ without using Laplace transforms?
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0@ Arturo Magidin: Thanks, I'm starting to use the forum and therefore need that kind of information. – 2011-11-07
4 Answers
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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0This is really amazing, thank you... – 2011-11-07
$\rm\bf Hint$: $\large e^{it}=\cos(t)+i\sin(t)\implies \sin(t)=\frac{e^{it}-e^{-it}}{2i}.$
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1@anon: So, $\sint$ = imaginary part of $e^{it}$ simplifies things better. – 2011-11-07
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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0thank you for finding the typo. – 2011-11-07
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.