Conditional convergence $\int_{-\infty}^{+\infty} \frac{(x-1)\sin 2x}{x^2-4x+5}dx$

Question

Explore conditional convergence $$\int_{-\infty}^{+\infty} \frac{(x-1)\sin 2x}{x^2-4x+5}dx$$ I tried $$\int_{-\infty}^{+\infty} \frac{(x-1)\sin 2x}{x^2-4x+5}dx = \int_{-\infty}^{+\infty} \frac{\sin 2x}{(x-1) - \frac{2(x-2)}{x-1}}dx=$$ $$\int_{-\infty}^{1} \frac{\sin 2x}{(x-1) - \frac{2(x-2)}{x-1}}dx + \int_{1}^{+\infty} \frac{\sin 2x}{(x-1) - \frac{2(x-2)}{x-1}}dx$$ but how to bound this? $$\int_{1}^{+\infty} \frac{\sin 2x}{(x-1) - \frac{2(x-2)}{x-1}}dx < \int_{1}^{+\infty} \frac{\sin 2x}{(x-1) - 3}dx$$ the last one diverges.

Answer 1

Dirichlet's test is enough. On $\mathbb{R}^+$ the function $\sin(2x)$ has a bounded primitive while $\frac{x-1}{x^2-4x+5}$ is eventually decreasing to zero. The integral on $\mathbb{R}^-$ is converging by the same argument.

The residue theorem also gives:

$$ \int_{-\infty}^{+\infty}\frac{(x-1)\sin(2x)}{(x-2)^2+1}\,dx = \color{red}{\frac{\pi}{e^2}\left(\sin 4+\cos 4\right)}.$$