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I want to calculate the integral below for positive $u$. $$\int_0^\infty \dfrac{t\sin{t}}{u^2+t^2}dt$$ Wolframalpha gives me a simple answer : $\dfrac{\pi}{2}e^{-u}$.

but I cannot approach to that. Can anyone solve above without using complex integral (ex.residue integration.. because I'm beginner of analysis)?

Thanks.

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    $\begingroup$ ${1 \over 2}\,\pi\exp\left(-\left\vert u\right\vert\right)$. $\endgroup$ – Felix Marin Nov 8 '16 at 3:49
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Hint. Let's consider the Laplace transform of $\displaystyle I(a):=\int_{0}^{\infty}\frac{\cos(ax)}{x^2+1}\:dx$. We have $$ \begin{aligned} \mathcal{L}\left(I(a)\right)(s)&=\mathcal{L}\left(\int_{0}^{\infty}\frac{\cos(ax)}{x^2+1}\:dx\right)(s) \\& = \int_{0}^{\infty}\int_{0}^{\infty}\frac{\cos(ax)}{x^2+1}e^{-as}\:da\:dx \\&= \int_{0}^{\infty}\frac{s}{(x^2+1)(s^2+x^2)}\;{dx} \\&= \frac{\pi}{2(s+1)} \end{aligned} $$giving $$ I(a)=\int_{0}^{\infty}\frac{\cos(ax)}{x^2+1}\:dx=\mathcal{L}^{-1}\left(\frac{\pi}{2(s+1)}\right) =\frac{\pi}{2}e^{-a},\qquad a>0, \tag2 $$ then by differentiating $(2)$ with respect to $a$, one gets $$ \int_{0}^{\infty}\frac{x\sin(ax)}{x^2+1}\:dx=\frac{\pi}{2}e^{-a},\qquad a>0. \tag3 $$ as given by Wolfram alpha.

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    $\begingroup$ Understand it, but how can I justify the exchange of improper integral between second and third equalities?? $\endgroup$ – Jinmu You Nov 7 '16 at 17:13
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    $\begingroup$ @JinmooYou This may be justified by the following inequalities: $\left|\int_{0}^{\infty}\frac{\cos(ax)}{x^2+1}\:dx \right|\le \int_{0}^{\infty}\frac1{x^2+1}\:dx<\infty$ and $ \left|\int_{0}^{\infty}\cos(ax)e^{-as}\:da\right|\le \int_{0}^{\infty}1\cdot e^{-as}\:da<\infty$ which give the right to use Fubini's theorem. $\endgroup$ – Olivier Oloa Nov 7 '16 at 19:52

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