Computing a real integral with complex integral method.

Question

I am to compute $\int_{-\infty}^{\infty}\frac{x\sin(3x)dx}{x^4+1}$.

So I geuss I should write the integral as $\int_{\tau}^{}\frac{z\sin(3z)dx}{z^4+1} -\lim_{R\rightarrow\infty}{\int_{C_{R}}^{}\frac{z\sin(3z)dx}{z^4+1}}$. Where is $\tau$ is half circle with the real line, and $C_R$ is without the line. $$\lim_{R\rightarrow\infty}{\int_{C_{R}}^{}\frac{z\sin(3z)dx}{z^4+1}}=0$$

So I am now left with $\int_{\tau}^{}\frac{z\sin(3z)dx}{z^4+1}$ and here I should find a residues of a function. What would be the best way here to do it, since it's the sine that's causing me difficulties.

Answer 1

Actually $\sin(z)$ diverges exponentially (for example approaching infinity along the imaginary axis), so the integral over the arc does not converge to zero and you cannot apply the usual ML estimate.

Instead

Hint: Write $\sin(x)=Im(e^{ix})$ and take the imaginary part outside the integral (which is legitimate as the function is real valued). Then calculate the new integral far right, using the usual half-circle contour:

$$\int_{-\infty}^{\infty}\frac{x\sin(3x)}{x^4+1}\,dx=\int_{-\infty}^{\infty}\frac{x\cdot Im(e^{3ix})}{x^4+1}\,dx=Im\left(\int_{-\infty}^{\infty}\frac{x\cdot e^{3ix}}{x^4+1}\,dx\right)$$