Question: How to evaluate this integral using residues∫∞0xsinx1+x2dx
I integrate over the entire real axis and dividing it by 1/2 since the integrand is even, and then I do the thing with the turning it into -\mathrm{Im}\bigg(\int f\cdot e^{ix} dx\bigg) and integrate over the upper half plane by finding the residues there, and there's only one, at i.
So I put it into the equation \frac{x e^{iz}}{2x} and get \frac{1}{2e}, so I get when plugging that in
\begin{align}\int_{0}^{\infty} \frac{x \sin x}{1 + x^2} dx &= .5 \int_{-\infty}^{\infty} \frac{x \sin x}{1 + x^2} dx \\&= .5 -\operatorname{Im} \Bigg[2\pi i \cdot \operatorname{Res}\bigg(\frac{x e^{ix}}{ 1 + x^2}; i\bigg)\Bigg]\\&= .5 \cdot \bigg[-\operatorname{Im}\Big(2\pi i\cdot \frac{1}{2e}\Big)\bigg] \\&= -\frac{\pi}{2e}\end{align}
But the answer's positive.
What did I mess up on?
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