# Integral over the real axis using residues

I have to evaluate the following integral $$\int_{-\infty}^\infty \frac{x\sin{x}}{x^2-4x+8}dx$$ I know that for such integrals we can use residues and $$\int_{-\infty}^\infty \frac{x\sin{x}}{x^2-4x+8}dx = 2\pi i \sum_{k=0}^n \text{res}[f(z),a_k] + \lim_{R\to\infty}\int_{\Gamma_R} f(z)dz$$ I see that $$f(z) = \frac{x\sin{x}}{x^2-4x+8}$$ has one singularity in the upper half of the plane which is $$x = 2+2i$$ and I have computed that res$$[f(z),2+2i] = \left(\frac{1}{2}-\frac{i}{2}\right)\sin{(2+2i)}$$, but I have problems evaluating the line integral. Usually we have shown that this goes to $$0$$ as $$R$$ gets large, but in this case it doesn't seem to be so. How can I go about evaluating the line integral? Thanks for any advice.

• the integral reduces to find the value of $\lim_{R\to \infty }\int_{\Gamma_R}\frac{\sin z}{z-\alpha }\,\mathrm d z$ where $\alpha$ is anyone of the roots of the polynomial. However I dont see a way to calculate it – Masacroso Nov 13 '19 at 16:20
• Change $\sin z$ to $e^{iz}$ in $f(z)$. – JanG Nov 13 '19 at 17:58

For integrals of the form $$\int_{-\infty}^{+\infty}R(x)e^{ix}dx$$ where $$R$$ is a rational function with a zero at infinity, consider a rectangle of vertexes \begin{align} &-X_1& &X_2& &X_2+iY& &-X_1+iY \end{align} for $$X_1,X_2,Y>0$$ large enough. Then we have \begin{align} &\left|\int_{X}^{X+iY}R(z)e^{iz}dz\right|\ll\frac 1X& &\left|\int_{-X_1+iY}^{X_2+iY}R(z)e^{iz}dz\right|\ll(X_1+X_2)\frac{e^{-Y}}Y \end{align}

This is a great question @MarkusPunnar. I am sad to see that no one gave a convincing answer.

According to my Complex Analysis book: To evaluate your integral we can examine the residues of the closely related $$\int_{-\infty}^\infty \frac{ze^{iz}}{z^2-4z+8}dz$$

The specific contour you wanted to look (the circular arc) at would be:

$$\lim_{R\to \infty} \int_0^\pi \frac{Re^{i\theta} e^{iRe^{i\theta}}}{R^2e^{i\theta}-4Re^{i\theta}+8}Re^{i\theta} d\theta$$

This expression is absurdly busy but it is cleaner than the expression you would get with $$sin(z).$$ The trick is to expand the $$e^{Re^{i\theta}}$$ term using Euler's formula. This will yield the ugly but approachable

$$\lim_{R\to \infty} \int_0^\pi \frac{R^2e^{2i\theta}e^{-Rsin(\theta)} e^{iRcos(\theta)}}{R^2e^{i\theta}-4Re^{i\theta}+8} d\theta$$

The modulus of this function will converge to $$0$$ as $$R\to\infty$$ on account of the $$e^{-Rsin(\theta)}$$ over the interval $$\theta \in (0,\pi)$$

$$\lim_{R\to \infty} | \frac{R^2e^{2i\theta}e^{-Rsin(\theta)} e^{iRcos(\theta)}}{R^2e^{i\theta}-4Re^{i\theta}+8} | = \lim_{R\to\infty}\frac{1}{e^{Rsin(\theta)}} = 0$$

Now we can use residues to establish that:

$$\int_{-\infty}^\infty \frac{ze^{iz}}{z^2-4z+8}dz = (\pi cos(2)- \pi sin(2))e^{-2} + i (\pi cos(2)+\pi sin(2))e^{-2}$$

By comparing imaginary parts we finally establish that:

$$\int_{-\infty}^\infty \frac{xsinx}{x^2-4x+8}dx = e^{-2}(\pi cos(2)+\pi sin(2))$$