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:

${\begin{aligned}&\int _{0}^{\infty }e^{-tx}\,dx=\left[{\frac {e^{-tx}}{-t}}\right]_{0}^{\infty }=\left(\lim _{x\to \infty }{\frac {e^{-tx}}{-t}}\right)-\left({\frac {e^{-t0}}{-t}}\right)\\&=0-\left({\frac {1}{-t}}\right)={\frac {1}{t}}.\end{aligned}}$

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

Failed to parse (unknown function "\begin{aligsn}"): {\displaystyle \int_0^\infty e^{-tx} \, dx = \frac{1}{t},</mathhhh> we can differentiate both sides twice with respect to ''t'' (not ''x'') in order to add the factor of ''x''<supp>2</sup> in the original integral. : <math> \begin{aligsn} & \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:

$\int _{0}^{\infty }x^{2}e^{-3x}\,dx={\frac {2}{3^{3}}}={\frac {2}{27}}.$