Integration by Parts

A first-order (single) integration by parts uses

$\begin{displaymath} d(uv) = u\,dv+v\,du \end{displaymath}$

(1)

$\begin{displaymath} \int d(uv) = uv = \int u\,dv + \int v\,du, \end{displaymath}$

(2)

$\begin{displaymath} \int u\,dv = uv - \int v\,du \end{displaymath}$

(3)

and

$\begin{displaymath} \int_a^b u\,dv = [uv]_a^b -\int_{f(a)}^{f(b)}v\,du. \end{displaymath}$

(4)

Now apply this procedure

times to $\int f^{(n)}(x)g(x)\,dx$ .

$\begin{displaymath} u=g(x)\qquad dv=f^{(n)}(x)\,dx \end{displaymath}$

(5)

$\begin{displaymath} du=g'(x)\,dx \qquad v=f^{(n-1)}(x). \end{displaymath}$

(6)

Therefore,

$\begin{displaymath} \int f^{(n)}g(x)\,dx = g(x)f^{(n-1)}(x)-\int f^{(n-1)}(x)g'(x)\,dx. \end{displaymath}$

(7)

But

$\begin{displaymath} \int f^{(n-1)}(x)g'(x)\,dx = g'(x)f^{(n-2)}(x)-\int f^{(n-2)}(x)g''(x)\,dx \end{displaymath}$

(8)

$\begin{displaymath} \int f^{(n-2)}(x)g''(x)\,dx = g''(x)f^{(n-3)}(x)-\int f^{(n-3)}(x)g^{(3)}(x)\,dx, \end{displaymath}$

(9)

$\int f^{(n)}(x)g(x)\,dx = g(x)f^{(n-1)}(x)-g'(x)f^{(n-2)}(x)$
$+g''(x)f^{(n-3)}(x)-\ldots +(-1)^n\int f(x)g^{(n)}(x)\,dx.\quad$	(10)

Now consider this in the slightly different form $\int f(x)g(x)\,dx$ . Integrate by parts a first time

$\begin{displaymath} u=f(x) \qquad dv=g(x)\,dx \end{displaymath}$

(11)

$\begin{displaymath} du=f'(x)\,dx \qquad v=\int g(x)\,dx, \end{displaymath}$

(12)

$\begin{displaymath} \int f(x)g(x)\,dx = f(x)\int g(x)\,dx -\int \left[{\int g(x)\,dx}\right]f'(x)\,dx. \end{displaymath}$

(13)

Now integrate by parts a second time,

$\begin{displaymath} u=f'(x) \qquad dv = \int g(x)\,dx \end{displaymath}$

(14)

$\begin{displaymath} du=f''(x)\,dx \qquad v=\int\!\!\!\int g(x)(dx)^2, \end{displaymath}$

(15)

$\displaystyle \int f(x)g(x)\,dx$	$\textstyle =$	$\displaystyle f(x)\int g(x)\,dx-f'(x)\int\!\!\!\int g(x)(dx)^2$
	$\textstyle \phantom{=}$	$\displaystyle + \int \left[{\int\!\!\!\int g(x)(dx)^2}\right]f''(x)\,dx.$	(16)

Repeating a third time,

$\int f(x)g(x)\,dx = f(x)\int g(x)\,dx-f'(x)\int\!\!\!\int g(x)(dx)^2+ f''(x)\int\!\!\!\int\!\!\!\int g(x)(dx)^3$

$- \int \left[{\int\!\!\!\int\!\!\!\int g(x)(dx)^3}\right]f'''(x)\,dx.\quad$ (17)

Therefore, after applications,

$\int f(x)g(x)\,dx = f(x)\int g(x)\,dx-f'(x)\int\!\!\!\int g(x)(dx)^2+ f''(x)\int\!\!\!\int\!\!\!\int g(x)(dx)^3-\ldots$

$+(-1)^{n+1}f^{(n)}(x) \underbrace{\int \cdots \int}_{n+1} g(x)(dx)^{n+1}$

$+(-1)^{n}\int\left[{\underbrace{\int \cdots \int\!\!}_{n+1} g(x)(dx)^{n+1}}\right]f^{(n+1)}(x)\,dx.\quad$ (18)

If $f^{(n+1)}(x)=0$ (e.g., for an th degree Polynomial), the last term is 0, so the sum terminates after terms and

$\int f(x)g(x)\,dx = f(x)\int g(x)\,dx$

$-f'(x)\int\!\!\!\int g(x)(dx)^2 + f''(x)\int\!\!\!\int\!\!\!\int g(x)(dx)^3-\ldots$

$+(-1)^{n+1}f^{(n)}(x) \underbrace{\int \cdots \int\!\!}_{n+1} g(x)(dx)^{n+1}.\quad$ (19)

References

Abramowitz, M. and Stegun, C. A. (Eds.). Handbook of Mathematical Functions with Formulas, Graphs, and Mathematical Tables, 9th printing. New York: Dover, p. 12, 1972.

$\int f(x)g(x)\,dx = f(x)\int g(x)\,dx-f'(x)\int\!\!\!\int g(x)(dx)^2+ f''(x)\int\!\!\!\int\!\!\!\int g(x)(dx)^3$
$- \int \left[{\int\!\!\!\int\!\!\!\int g(x)(dx)^3}\right]f'''(x)\,dx.\quad$	(17)

$\int f(x)g(x)\,dx = f(x)\int g(x)\,dx-f'(x)\int\!\!\!\int g(x)(dx)^2+ f''(x)\int\!\!\!\int\!\!\!\int g(x)(dx)^3-\ldots$
$+(-1)^{n+1}f^{(n)}(x) \underbrace{\int \cdots \int}_{n+1} g(x)(dx)^{n+1}$
$+(-1)^{n}\int\left[{\underbrace{\int \cdots \int\!\!}_{n+1} g(x)(dx)^{n+1}}\right]f^{(n+1)}(x)\,dx.\quad$	(18)

$\int f(x)g(x)\,dx = f(x)\int g(x)\,dx$
$-f'(x)\int\!\!\!\int g(x)(dx)^2 + f''(x)\int\!\!\!\int\!\!\!\int g(x)(dx)^3-\ldots$
$+(-1)^{n+1}f^{(n)}(x) \underbrace{\int \cdots \int\!\!}_{n+1} g(x)(dx)^{n+1}.\quad$	(19)