# Integration using parametric derivatives

In mathematics, integration by parametric derivatives is a method of integrating certain functions.

For example, suppose we want to find the integral

$\int_0^\infty x^2 e^{-3x} \, dx.$

Since this is a product of two functions that are simple to integrate separately, repeated integration by parts is certainly one way to evaluate it. However, we may also evaluate this by starting with a simpler integral and an added parameter, which in this case is t = 3:

\begin{align} & \int_0^\infty e^{-tx} \, dx = \left[ \frac{e^{-tx}}{-t} \right]_0^\infty = \left( \lim_{x \to \infty} \frac{e^{-tx}}{-t} \right) - \left( \frac{e^{-t0}}{-t} \right) \\ & = 0 - \left( \frac{1}{-t} \right) = \frac{1}{t}. \end{align}

This converges only for t > 0, which is true of the desired integral. Now that we know

$\int_0^\infty e^{-tx} \, dx = \frac{1}{t},$

we can differentiate both sides twice with respect to t (not x) in order to add the factor of x2 in the original integral.

\begin{align} & \frac{d^2}{dt^2} \int_0^\infty e^{-tx} \, dx = \frac{d^2}{dt^2} \frac{1}{t} \\[10pt] & \int_0^\infty \frac{d^2}{dt^2} e^{-tx} \, dx = \frac{d^2}{dt^2} \frac{1}{t} \\[10pt] & \int_0^\infty \frac{d}{dt} \left (-x e^{-tx}\right) \, dx = \frac{d}{dt} \left(-\frac{1}{t^2}\right) \\[10pt] & \int_0^\infty x^2 e^{-tx} \, dx = \frac{2}{t^3}. \end{align}

This is the same form as the desired integral, where t = 3. Substituting that into the above equation gives the value:

$\int_0^\infty x^2 e^{-3x} \, dx = \frac{2}{3^3} = \frac{2}{27}.$