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Evaluate the definite integral $$\int_0^\infty \frac{x\sin mx}{x^2+a^2}dx \quad (m,a>0)$$

I tried a trigonometric substitution but did not get anywhere with that I think the multiple variable are throwing me off.

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That integral is screaming please solve me through the residue theorem.

For first, we get rid of one parameter by substituting $x=az$. Then we have to compute: $$ I(k) = \int_{0}^{+\infty}\frac{x}{x^2+1}\cdot\sin(kx)\,dx=\frac{1}{2}\int_{\mathbb{R}}\frac{x}{x^2+1}\cdot\sin(kx)\,dx $$ that is half the imaginary part of $\int_{\mathbb{R}}\frac{x e^{ikx}}{x^2+1}\,dx$. By computing the residue of the integrand function at $x=i$ it follows that: $$ I(k) = \frac{\pi}{2}\cdot e^{-k}, $$ hence:

$$\int_{0}^{+\infty}\frac{x\sin(mx)}{x^2+a^2}\,dx = \color{red}{\frac{\pi}{2}\cdot e^{-am}}.$$

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  • $\begingroup$ I don't see the integral along the arc of the contour going to zero... $\endgroup$ – T.J. Gaffney Apr 7 '16 at 17:02
  • $\begingroup$ @Gaffney: the integration path should be a rectangle enclosing $x=i$, indeed. $\endgroup$ – Jack D'Aurizio Apr 7 '16 at 17:11
  • $\begingroup$ The top part of the rectangle would be zero? $\endgroup$ – T.J. Gaffney Apr 7 '16 at 17:35
  • $\begingroup$ @Gaffney: the situation is almost the same as the usual proof of $\int_{0}^{+\infty}\frac{\sin x}{x}\,dx=\frac{\pi}{2}$. Have a look at robjohn's proof here: math.stackexchange.com/questions/594641/… $\endgroup$ – Jack D'Aurizio Apr 7 '16 at 17:39

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