A way of evaluating integrals without doing anything?

Solution 1:

Rather than try and address this question generally here we give another---hopefully sufficiently different---example of the use of this type of machinery.

Suppose we are interested in integrals of the form $$\begin{equation*} \int_0^{2\pi} x^n \sin x\, dx \quad\textrm{ or }\quad \int_0^{2\pi} x^n \cos x \, dx, \end{equation*}$$ where $n\in\mathbb{N}$. It is natural to consider the more general integral $$\begin{equation*} I_n = i\int_0^{2\pi} x^n e^{i x}\, dx. \end{equation*}$$ (A factor of $i$ has been introduced for convenience.) Then $$\begin{eqnarray*} I(r) &\equiv& \sum_{n=0}^\infty \frac{(i r)^n}{n!}I_n \\ &=& i\int_0^{2\pi} e^{i (r+1) x}dx \\ &=& \frac{e^{2\pi i r}-1}{1+r} \\ &=& \sum_{n=0}^\infty \left(\sum_{k=0}^{n-1}(-1)^{k}\frac{(2\pi i)^{n-k}}{(n-k)!}\right)r^n. \end{eqnarray*}$$ (The last series is the Cauchy product of the Taylor series for $1/(1+r)$ and the Taylor series for $e^{2\pi i r}-1$.) Therefore, $$\begin{equation*} \int_0^{2\pi} x^n e^{i x}\, dx = \frac{n!}{i^{n+1}}\sum_{k=0}^{n-1} (-1)^{k} \frac{(2\pi i)^{n-k}}{(n-k)!}.\tag{1} \end{equation*}$$ By taking the real and imaginary part of (1) we can find explicit formulas for the original integrals: $$\begin{eqnarray*} \int_0^{2\pi} x^n \cos x\, dx &=& n!\sum_{m=0}^{\lfloor\frac{n-2}{2}\rfloor} (-1)^m\frac{(2\pi)^{n-2m-1}}{(n-2m-1)!} \\ \int_0^{2\pi} x^n \sin x\, dx &=& n!\sum_{m=0}^{\lfloor\frac{n-1}{2}\rfloor} (-1)^{m+1}\frac{(2\pi)^{n-2m}}{(n-2m)!}. \end{eqnarray*}$$

Solution 2:

Nothing magic :-) (if I understand the question correctly.) If $f(r)$ is an analytic function around $r_0=0$ then the Taylor expansion is unique. It yields, $$ f(r)=\sum_{n=1}^{\infty}c_nr^n, $$ where $r\in(-a,a)$, $a>0$. So, if $$ f(r)=\sum_{n=1}^{\infty}c_n d_n r^n $$ is another power series representation of $f$, then $c_n=c_n d_n$, i.e., $d_n=1$ if $c_n\neq 0$, anything is $d_n$ (integral or something else).

Solution 3:

$$\sum_{n=0}^\infty \frac{r^n}{n!} \int_0^\infty x^n e^{-x} \; dx = \sum_{n=0}^\infty \frac{r^n}{n!} \left(\int_0^\infty x^n e^{-x} \; dx\right) = \sum_{n=0}^\infty \frac{r^n}{n!} \Gamma(n+1) = \sum_{n=0}^\infty \frac{r^n}{n!} n! = \sum_{n=0}^\infty r^n$$

Where $\Gamma(n+1)$ is the Gamma function. I hope this is right!