Improper integral : $\int_0^{+\infty}\frac{x\sin x}{x^2+1}$ [closed]

Solution 1:

Since your integrand is even, this integral is equal to one half of $$ \int _{-\infty}^{\infty}\frac{x\sin x}{x^2+1}dx $$

This integral is the imaginary part of

$$ \int _{-\infty}^{\infty}\frac{x \cdot e^{ix}}{x^2+1}dx $$

This can be solved as the contour integral with contour a half disc of radius $R$ with base on the real axis, letting $R$ go to infinity. The integral along the arc goes to $0$ ( needs some showing ). The contour integral is equal to the residue at $x=i$, which is $e^{-1}\pi i$. So taking half the imaginary part, we get $\frac{\pi}{2e}$.

Solution 2:

Using the result from this OP: Integral evaluation $\int_{-\infty}^{\infty}\frac{\cos (ax)}{\pi (1+x^2)}dx$. We have \begin{equation} \int_0^\infty\frac{\cos ax}{1+x^2}\,dx=\frac{\pi}{2}e^{-|a|} \end{equation} Thus, our integration is simply \begin{align} \int_0^\infty\frac{x\sin x}{1+x^2}\,dx&=-\lim_{a\to1}\partial_a\int_0^\infty\frac{\cos ax}{1+x^2}\,dx\\ &=-\frac{\pi}{2}\lim_{a\to1}\partial_a\left[e^{-|a|}\right]\\ &=\frac{\pi}{2e} \end{align}

Solution 3:

According to the residue theory,

$$ \int_0^{+\infty}\frac{1}{s^2+x^2}\mathrm{d}x=\frac{\pi}{2s} ~ , ~ I(\alpha)=\int_0^{+\infty}\frac{x\sin \alpha x}{1+x^2}\mathrm{d}x $$

Laplace transform:

\begin{align} \mathcal{L}\left[ I(\alpha)\right] &= \int_0^{+\infty}\frac{x}{1+x^2}\cdot \frac{x}{s^2+x^2}\mathrm{d}x \\ &= \int_0^{+\infty}\frac{x^2+1-1}{1+x^2} \cdot \frac{1}{s^2+x^2} \mathrm{d}x \\ &= \int_0^{+\infty}\frac{1}{s^2+x^2} \mathrm{d}x - \int_0^{+\infty}\frac{1}{1+x^2} \cdot \frac{1}{s^2+x^2} \mathrm{d}x \\ &= \int_0^{+\infty}\frac{1}{s^2+x^2} \mathrm{d}x - \frac{1}{s^2-1}\int_0^{+\infty}\left( \frac{1}{1+x^2} - \frac{1}{s^2+x^2} \right) \mathrm{d}x \\ &= \frac{\pi}{2} \cdot \frac{1}{s+1} \end{align}

Inverse transform:

$$\mathcal{L}^{-1}\left[ I(\alpha)\right] = \frac{\pi}{2}e^{-\alpha} \Longrightarrow I(1)=\int_0^{+\infty}\frac{x\sin x}{1+x^2} \mathrm{d}x=\frac{\pi}{2e}$$