Striking applications of integration by parts

Solution 1:

I always liked the derivation of Taylor's formula with error term:

$$\begin{array}{rl} f(x) &= f(0) + \int_0^x f'(x-t) \,dt\\ &= f(0) + xf'(0) + \int_0^x tf''(x-t)\,dt\\ &= f(0) + xf'(0) + \frac{x^2}2f''(0) + \int_0^x \frac{t^2}2 f'''(x-t)\,dt \end{array}$$

and so on. Using the mean value theorem on the final term readily gives the Cauchy form for the remainder.

Solution 2:

Let $f$ be a differentiable one-to-one function, and let $f^{-1}$ be its inverse. Then,

$$\int f(x) dx = x f(x) - \int x f'(x)dx = x f(x) - \int f^{-1}(f(x))f'(x)dx = x f(x) - \int f^{-1}(u) du \,.$$

Thus, if we know the integral of $f^{-1}$, we get the integral of $f$ for free.

BTW: This is the reason why the integrals $\int \ln(x) dx \,;\, \int \arctan(x) dx \,; ...$ are always calculated using integration by parts.

Solution 3:

My favorite this week, since I learned it just yesterday: $n$ integrations by parts produces $$ \int_0^1 \frac{(-x\log x)^n}{n!}dx = (n+1)^{-(n+1)}.$$ Then summing on $n$ yields $$\int_0^1 x^{-x}\,dx = \sum_{n=1}^\infty n^{-n}.$$

Solution 4:

Repeated integration by parts gives $$\int_0^\infty x^n e^{-x} dx=n!$$