Why is $\int e^{-xy}\sin{x} dx = \frac{e^{-xy}(y\sin{x}+\cos{x})}{1+y^2}$

Question

Why would $\int e^{-xy}\sin{x} dx = \frac{e^{-xy}(y\sin{x}+\cos{x})}{1+y^2}$

My calculation:

$\int e^{-xy}\sin{x} dx=\Im\int e^{-xy}(\cos{x}+i\sin{x})dx$

and $\int e^{-xy}(\cos{x}+i\sin{x})dx=-\frac{1}{y-i}e^{x(i-y)}=-\frac{1}{y^2+1}e^{x(i-y)}$

Now $\Im -\frac{1}{y^2+1}e^{x(i-y)}=\Im-\frac{e^{-xy}}{y^{2}+1}(\cos{x}+i\sin{x})=-\frac{e^{-xy}}{y^{2}+1}\sin{x}$

In the solutions, it says: $\int e^{-xy}\sin{x} dx=\frac{e^{-xy}(y\sin{x}+\cos{x})}{1+y^2}$

Where am I going wrong?

Answer 1

You dropped a $y+i$ factor before "now".

Answer 2

I get $$\int e^{-xy}\sin{x}\; dx = \color{red}{-}\frac{e^{-xy}(y\sin{x}+\cos{x})}{1+y^2} (+C)$$ Here a way using partial integration twice:

With $I(y) = \int \underbrace{e^{-xy}}_{u}\underbrace{\sin{x}}_{v'}\;dx$ you get \begin{eqnarray*} I(y) & = & -e^{-xy}\cos x - y \int e^{-xy}\cos{x} \; dx \\ & = & -e^{-xy}\cos x - y \left(e^{-xy} \sin x + y \underbrace{\int e^{-xy}\sin{x} \; dx }_{= I(y)}\right)\\ & = & -e^{-xy}\cos x - ye^{-xy} \sin x - y^2 I(y)\\ \end{eqnarray*} It follows $$(1+y^2)I(y) = - -e^{-xy}(\cos x +y \sin x) \Leftrightarrow \boxed{I(y) = -\frac{e^{-xy}(y\sin{x}+\cos{x})}{1+y^2} (+C)}$$