info prev up next book cdrom email home

Integration by Parts

A first-order (single) integration by parts uses

d(uv) = u\,dv+v\,du
\end{displaymath} (1)

\int d(uv) = uv = \int u\,dv + \int v\,du,
\end{displaymath} (2)

\int u\,dv = uv - \int v\,du
\end{displaymath} (3)

\int_a^b u\,dv = [uv]_a^b -\int_{f(a)}^{f(b)}v\,du.
\end{displaymath} (4)

Now apply this procedure $n$ times to $\int f^{(n)}(x)g(x)\,dx$.
u=g(x)\qquad dv=f^{(n)}(x)\,dx
\end{displaymath} (5)

du=g'(x)\,dx \qquad v=f^{(n-1)}(x).
\end{displaymath} (6)

\int f^{(n)}g(x)\,dx = g(x)f^{(n-1)}(x)-\int f^{(n-1)}(x)g'(x)\,dx.
\end{displaymath} (7)


\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)

\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
u=f(x) \qquad dv=g(x)\,dx
\end{displaymath} (11)

du=f'(x)\,dx \qquad v=\int g(x)\,dx,
\end{displaymath} (12)


\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,
u=f'(x) \qquad dv = \int g(x)\,dx
\end{displaymath} (14)

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 $n$ 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 $n$th degree Polynomial), the last term is 0, so the sum terminates after $n$ 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)


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.

info prev up next book cdrom email home

© 1996-9 Eric W. Weisstein