Derivative of product

Suppose $u=u(x), v=v(x)$. By definition of derivative:

\[\begin{align} (uv)' &= \lim_{\delta \rightarrow 0} \frac{u(x + \delta) v(x + \delta) - u(x)v(x)}{\delta} \\ &= \lim_{\delta \rightarrow 0} \left[ \frac{u(x + \delta) v(x + \delta) - u(x)v(x + \delta)}{\delta} + \frac{u(x)v(x + \delta) - u(x)v(x)}{\delta} \right ] \label{eq:minus_plus_trick} \\ &= \lim_{\delta \rightarrow 0} \left [ \frac{u(x + \delta) v(x + \delta) - u(x)v(x + \delta)}{\delta} + \frac{u(x)v(x + \delta) - u(x)v(x)}{\delta} \right ] \\ &= \lim_{\delta \rightarrow 0} \left [\frac{(u(x + \delta) - u(x)) v(x + \delta)}{\delta} + \frac{u(x)(v(x + \delta) - v(x))}{\delta} \right ] \\ &= u'v + uv' \end{align}\]

Note in Eq. $\eqref{eq:minus_plus_trick}$, we simply minus $u(x)v(x + \delta)$ and then added it back in the numerator.

Therefore, we have the product rule:

\[\begin{align} (uv)' = u'v + uv' \label{eq:product_rule} \end{align}\]

Integration by parts

Eq. $\eqref{eq:product_rule}$ can be rearranged into

\[\begin{align} uv' &= (uv)' - u'v \\ \int uv' dx &= \int (uv)' dx - \int u'v dx \label{eq:integration_by_parts} \\ \end{align}\]

Eq. $\eqref{eq:integration_by_parts}$ is usally referred to as integration by parts. It is useful when integrating the product of two functions, e.g. $a(x) = u, b(x) = v$, esp. $uv’$ is hard to integrate, but $(uv)’$ and $u’v$ is easy to integrate. We’ll demonstrate such a use case with the proof for the factorial property of the Gamma function.

Gamma function generalizes factorial

The gamma function is defined as

\[\begin{align} \Gamma(n) &=\int_0^{\infty} x^{n - 1} e^{-x} dx \end{align}\]

Note, the parameter for $\Gamma(n)$ is $n$, not $x$, which is just a dummy variable that will be integrated out. $\Gamma(n)$ is a generalization fo the factorial function $n!$, and it has the factorial property

\[\begin{align} \Gamma(n + 1) = n \Gamma(n) \end{align}\]

Let’s prove it. Note, integrating

\[\begin{align} \Gamma(n + 1) = \int_0^{\infty} x^n e^{-x} dx \end{align}\]

is hard, but if we consider

\[\begin{align} u &= x^n \\ v' &= e^{-x} \\ \end{align}\]

then, we have

\[\begin{align} u' &= n x^{n - 1} \\ v &= - e^{-x} \\ \end{align}\]

Then, applying integration by parts (Eq. $\eqref{eq:integration_by_parts}$), both $uv$ and $u’v$ are easy to integrate.

\[\begin{align} \Gamma(n + 1) &= \int_{0}^{\infty} x^n e^{-x} dx \\ &= \left[ - x^n e^{-x} \right ]_{0}^{\infty} + n \int_0^{\infty} x^{n-1} e^{-x} dx \\ &= 0 + n \Gamma(n) \\ &= n \Gamma(n) \\ \end{align}\]

Alternatively, given $x^n$ is equally easy to integrate, we can also set

\[\begin{align} u &= e^{- x} \\ v' &= x^n \\ u' &= - e^{- x} \\ v &= \frac{1}{n + 1} x^{n + 1} \end{align}\]

and

\[\begin{align} \Gamma(n+1) &= \int_{0}^{\infty} x^n e^x dx \\ &= \left[ - x^n e^x \right ]_{0}^{\infty} + \frac{1}{n+1} \int_0^{\infty} x^{n+1} e^x dx \\ (n + 1) \Gamma(n+1) &= 0 + \Gamma(n + 2) \\ \Gamma(n + 2) &= (n + 1) \Gamma(n+1) \\ \end{align}\]

which is equivalent to $\Gamma(n) = n\Gamma(n - 1)$.