Euler Complexification Help.

Question

We have$$e^{ix}=\cos x+i\sin x$$

I want

$$\begin{align} I&=\large\int \underbrace{\sin x}_{\large\Im e^{ix}}\quad e^{\pi x}\,\mathrm dx \\ &=\int e^{ix}e^{\pi x}dx\\ &=\int e^{(i+\pi) x}dx\\ &=\frac{e^{(i+\pi) x}}{(i+\pi)}\\ &=\frac{\pi-i}{\pi^2+1}e^{\pi x}(\cos x +i\sin x)\\ &=\color{blue}{\frac{1}{\pi^2+1}e^{\pi x}(\pi-i)(\cos x +i\sin x)}\\ &I=\color{blue}{\Im\left(\frac{1}{\pi^2+1}e^{\pi x}(\pi\sin x-\cos)\right)}\\ \end{align}$$

Please explain how my teacher distributed the Pi-i (circled in blue).

Answer 1

We have

$$\begin{align} I &=\int e^{\pi x}\sin x\,\mathrm dx\\ &=\Im\int e^{(\pi+i) x}\,\mathrm dx\\ &=\color{blue}{\Im}\frac{e^{\pi x}}{\pi^2+1}(\pi-i)(\cos x+i\sin x)\\ &=\color{blue}{\Im}\frac{e^{\pi x}}{\pi^2+1}\left({\pi\cos x}+i\color{blue}{\pi\sin x}-i\color{blue}{\cos x}+{\sin x}\right)\\ &=\frac{e^{\pi x}}{\pi^2+1}\left(\color{blue}{\pi\sin x}-\color{blue}{\cos x}\right)\\ \end{align}$$

$$\large \int e^{\pi x}\sin x\,\mathrm dx=\frac{e^{\pi x}}{\pi^2+1}\left({\pi\sin x}-{\cos x}\right)$$