Integration by parts loop with coefficient

Question

I leant that when the integral appears on the right side of the equation, it can be transferred accross to the left side, as in this post, but I'm trying to learn how to do this if there is a coefficient with the integral on the RHS. For example:

\begin{align}\int e^{-int}\sin(t)\,dt&=\frac{i}{n}e^{-int}\sin(t)-\int\frac{i}{n}e^{-int}\cos(t)\,dt \\ &=\frac{i}{n}e^{-int}\sin(t)-\left(-\frac{1}{n^2}e^{-int}\cos(t)-\int\frac{1}{n^2}e^{-int}\sin(t)\,dt\right)\end{align}

I can see that the integral appears on the RHS, and normally this could be transferred across to the LHS, but I'm not sure of how it all works with the $\frac{1}{n^2}$

Any help or direction would be much appreciated.

Answer 1

You have $$ \int e^{-int} \sin(t) \, dt = \frac{i}{n}e^{-int}\sin(t) + \frac{1}{n^2}e^{-int}\cos(t) + \frac{1}{n^2} \int e^{-int}\sin(t) $$

So, $$ \left( 1 - \frac{1}{n^2} \right) \int e^{-int} \sin(t) \, dt = \frac{i}{n}e^{-int}\sin(t) + \frac{1}{n^2}e^{-int}\cos(t) $$ i.e. $$ \int e^{-int} \sin(t) \, dt = \left( 1 - \frac{1}{n^2} \right)^{-1} \left( \frac{i}{n}e^{-int}\sin(t) + \frac{1}{n^2}e^{-int}\cos(t) \right) \\ = \frac{n^2}{n^2-1} \left( \frac{i}{n}e^{-int}\sin(t) + \frac{1}{n^2}e^{-int}\cos(t) \right) \\ = \frac{in}{n^2-1}e^{-int}\sin(t) + \frac{1}{n^2-1}e^{-int}\cos(t) \\ $$

Answer 2

If you denote $\color{#df0000}{I := \int e^{- i n t} \sin t \,dt}$, the equation becomes $$\color{#df0000}{I} = \frac{1}{n} e^{-int} \sin t + \frac{1}{n^2} e^{-int} \cos t + \frac{1}{n^2} \color{#df0000}{I} + C$$ for a constant $C$ of integration. Now, solve for $\color{#df0000}{I}$, which we can do iff $n \neq \pm 1$.

Remark Alternatively you can evaluate this integral by writing $\sin t = \frac{1}{2 i} (e^{i t} - e^{-i t})$ and distributing to get a linear combination of the integrals $\int e^{-i(n \mp 1)t} dt$---this formulation makes it clearer why the cases $n = \pm 1$ are qualitatively different from the case of generic $n$.