May 1, 2023

By definition, $\displaystyle\int f g\;dx$ can be expressed as

$\displaystyle \int f \left(\int g\; dx \right)' \;dx.$

Applying integration by parts,

$\displaystyle\int fg\;dx$

$\overset{g_1=\int g\;dx }{=} \displaystyle\int f g_1'\;dx = f g_1 - \underline{\int f' g_1\;dx}$

$\overset{g_2 = \int g_1\;dx}{=} f g_1 -\left(f' g_2 -\displaystyle \int f'' g_2\;dx\right)$

$= f g_1 -f' g_2 + \displaystyle \int f'' g_2\;dx$

$\overset{g_3 = \int g_2\;dx}{=} f g_1 -f' g_2 + \left(f'' g_3 - \displaystyle\int f^{(3)} g_3\;dx\right)$

$\ddots$

$= fg_1 - f'g_2 + f^{''}g_3 + \dots + (-1)^{n-1} f^{(n-1)} g_n +(-1)^n\displaystyle\int f^{n}g_n\;dx$

$= \sum\limits_{i=1}^{n}(-1)^{i-1}f^{i-1}g_i + (-1)^n\displaystyle\int f^{(n)}g_n\;dx$

Note a pattern has emerged – namely,

$\displaystyle \int fg\;dx = fg_1 - f'g_2 + f^{''}g_3 + \dots + (-1)^{n-1} f^{(n-1)} g_n +(-1)^n\displaystyle\int f^{n}g_n\;dx$

holds for all $n$ ‘s (see $(*)$ ). Consequently, it introduces a schematic method of evaluating certain integrals. For example, to determine

$I_1 = \displaystyle \int x^5 \sin(x)\;dx,$

let $f=x^5$ and $g=\sin(x).$ Write $f'$ in one column and $\displaystyle\int g\; dx$ in another, continuing to the row in which $f^{(n)} = 0:$

$\displaystyle I_1=$ $-x^5\cos(x) + 5x^4\sin(x)+20x^3\cos(x)-60x^2\sin(x)-120x\cos(x)+120\sin(x)$

To evaluate

$I_2 = \displaystyle\int e^{ax}\cos(bx)\;dx,$

let $f = e^{ax}$ and $g=\sin(bx).$ Proceed as before, but continuing to the row in which the $f_n g_n$ is a constant multiple of $I_2$ ‘s integrand:

$I_2 = e^{ax}\cdot\frac{\sin(bx)}{b} + ae^{ax}\cdot\frac{cos(bx)}{b^2} - \frac{a^2}{b^2}I_2.$

Solving for $I_2,$ we obtain

$I_2 = \frac{e^{ax}(b\cdot\sin(bx)+a\cdot\cos(bx))}{a^2+b^2}$

Prove:

$\forall n \in \mathbb{N}, \displaystyle \int fg\;dx = \sum\limits_{i=1}^{n}(-1)^{i-1}f^{i-1}g_i + (-1)^n\displaystyle\int f^{(n)}g_n\;dx\quad\quad\quad(*)$

where $f^{(0)} = f, f^{(n)} = (f^{(n-1)})'; g_{0} = g, g_{n} = \displaystyle\int g_{n-1}\;dx.$

proof

When $n=1,$ the right-hand side of $(*)$ is

$\sum\limits_{i=1}^{1}(-1)^{i-1}f^{i-1}g_i + (-1)^1\displaystyle\int f^{(1)}g_1\;dx$

$= f^{(0)} g_1 - \displaystyle\int f^{(1)}g_1\;dx$

$= f\displaystyle\int g_0\;dx - \int f' \left(\int g_0\;dx \right)\;dx$

$= \displaystyle\int f \left(\int g_0\;dx\right)'\;dx$

$= \displaystyle\int f g\;dx$

which is the left-hand side of $(*).$

Suppose when $n = k,$

$\displaystyle\int f\cdot g\;dx = \sum\limits_{i=1}^{k}(-1)^{i-1}f^{i-1}g_i + (-1)^k\displaystyle\int f^{(k)}g_k\;dx.$

Then

$\displaystyle\int f\cdot g\;dx = \sum\limits_{i=1}^{k}(-1)^{i-1} f^{(i-1)}g_i +(-1)^k\displaystyle\int f^{(k)}(g_{k+1})'\; dx$

$= \sum\limits_{i=1}^{k}(-1)^{i-1} f^{(i-1)}g_i +(-1)^k \left(f^{(k)}g_{k+1} - \displaystyle\int f^{(k+1)}g_{k+1}\;dx\right)$

$= \sum\limits_{i=1}^{k}(-1)^{i-1} f^{(i-1)}g_i +(-1)^k f^{(k)}g_{k+1} +(-1)^{k+1}\displaystyle\int f^{(k+1)}g_{k+1}\;dx$

$=\underbrace{\sum\limits_{i=1}^{k}(-1)^{i-1} f^{(i-1)}g_i +(-1)^{(k+1-1)} f^{(k+1-1)}g_{k+1}}_{\sum\limits_{i=1}^{k+1}(-1)^{i-1} f^{(i-1)}g_i} +(-1)^{k+1}\displaystyle\int f^{(k+1)}g_{k+1}\;dx$