Calculating improper complex integral $\int_0^{+\infty}(x^2+2x)\exp(\omega x)\sin(x) \mathrm{dx}$

Question

$$\int_0^{+\infty}(x^2+2x)\exp(\omega x)\sin(x) \mathrm{dx}$$

$$\omega \in \mathbb{C}, \mathrm{Re}(\omega)<0$$

How do I solve this - what's the best way to calculate this integral? Looking at real part and imaginary part separately?

Answer 1

First, let's rename the variable $\omega =-s$ such that we obtain

$$\int_0^{+\infty}(x^2+2x)\exp(-s x)\sin(x) \mathrm{dx}.$$

Then use $s=a+jb$ and Euler's formula as a substitution to obtain

$$\int_0^{+\infty}(x^2+2x)\exp(-a x-jbx)\sin(x) \mathrm{dx}=\int_0^{+\infty}(x^2+2x)\exp(-a x)\exp(-jbx)\sin(x) \mathrm{dx}$$ $$=\int_0^{+\infty}(x^2+2x)\exp(-a x)\left[\cos(bx)-j\sin(bx)\right]\sin(x) \mathrm{dx}$$ $$=\int_0^{+\infty}(x^2+2x)\exp(-a x)\cos(bx)\sin(x) \mathrm{dx}$$ $$-j\int_0^{+\infty}(x^2+2x)\exp(-a x)\sin(bx)\sin(x) \mathrm{dx}.$$

In the next step apply

$$\sin u \cos v= \dfrac{1}{2}\left[\sin(u-v)+\sin(u+v) \right]$$ $$\sin u \sin v= \dfrac{1}{2}\left[\cos(u-v)+\cos(u+v) \right]$$

to turn the products of trigonometric functions into sums of trigonometric functions.

Then apply Euler's integral formulas to obtain the solutions to the appearing integrals.

Answer 2

$\newcommand{\bbx}[1]{\,\bbox[15px,border:1px groove navy]{\displaystyle{#1}}\,} \newcommand{\braces}[1]{\left\lbrace\,{#1}\,\right\rbrace} \newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack} \newcommand{\dd}{\mathrm{d}} \newcommand{\ds}[1]{\displaystyle{#1}} \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,} \newcommand{\ic}{\mathrm{i}} \newcommand{\mc}[1]{\mathcal{#1}} \newcommand{\mrm}[1]{\mathrm{#1}} \newcommand{\pars}[1]{\left(\,{#1}\,\right)} \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,} \newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}} \newcommand{\verts}[1]{\left\vert\,{#1}\,\right\vert}$ \begin{align} &\bbox[10px,#ffd]{\ds{\left.\int_{0}^{\infty}\pars{x^{2} + 2x} \exp\pars{\omega x}\sin\pars{x}\,\dd x\,\right\vert_{\ \Re\pars{\omega}\ <\ 0}}} \\[5mm] = &\ \pars{\partiald[2]{}{\omega} + 2\,\partiald{}{\omega}} \int_{0}^{\infty}\exp\pars{\omega x}\,{\expo{\ic x} - \expo{-\ic x} \over 2\ic}\,\dd x = -\,{1 \over 2}\,\ic\pars{\partiald[2]{}{\omega} + 2\,\partiald{}{\omega}} \pars{{-1 \over \omega + \ic} - {-1 \over \omega - \ic}} \\[5mm] = &\ -\,{1 \over 2}\,\ic\pars{\partiald{}{\omega} + 2} \bracks{{1 \over \pars{\omega + \ic}^{2}} - {1 \over \pars{\omega - \ic}^{2}}} \\[5mm] = &\ -\,{1 \over 2}\,\ic \bracks{-\,{2 \over \pars{\omega + \ic}^{3}} + {2 \over \pars{\omega - \ic}^{3}} + {2 \over \pars{\omega + \ic}^{2}} - {2 \over \pars{\omega - \ic}^{2}}} \\[5mm] = &\ \bbx{{\ic \over \pars{\omega + \ic}^{3}} - {\ic \over \pars{\omega - \ic}^{3}} - {\ic \over \pars{\omega + \ic}^{2}} + {\ic \over \pars{\omega - \ic}^{2}}} \end{align}