Why is infinite Integration by Parts valid?

Question

Few days ago I wrote an answer to solving $\int x\exp(x^2)$ using integration by parts,the general formula for the integral is $$\int x^{2n+1}e^{x^2}dx=\frac{x^{2n+2}}{2n+2}e^{x^2}-\frac{1}{n+1}\int x^{2n+3}e^{x^2}dx$$ If we label the integral $I_n$ we get $$I_n=\frac{x^{2n+2}}{2n+1}e^{x^2}-\frac{1}{n+1}I_{n+1}$$ Now my question is why can we substitute the integral for an infinite sum,like this (for clarity lets omit $C$)? $$I_0=\frac{x^2}{2}e^{x^2}-\frac{x^4}{4}e^{x^2}+\frac{x^6}{12}e^{x^2}-\frac{x^8}{48}e^{x^2}+\frac{x^{10}}{240}e^{x^2}+\cdots$$ It seems like we are omitting the $I_{n+1}$ somehow, what is the correct justification for this step?

Answer 1

To ask a precise question, consider the definite integral over an interval $[a,b]$.

Recall that a series is just a sequence of partial sums. And if $A_j$ is the general term of the series that you get in the end, you have something like:

$$ I_0 = \sum_{j=1}^n A_j - \frac{1}{(n+1)!}I_{n+1} $$

And so the partial sums converge precisely when $\frac{1}{n!}I_n$ converges. Now you can try to bound $$\frac{1}{n!}I_{n} = \frac{1}{n!}\int_a^b x^{2n+1}e^{x^2}dx$$ to show that the series converges for any $a$, $b$.