Integration using parametric derivatives

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

For an example, suppose we want to find the integral

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

Seeing this as 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 only converges for $t>0$, which is true for 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 $x^{2}$ in the original integral.

${\frac {d^{2}}{dt^{2}}}\int _{0}^{\infty }e^{-tx}dx={\frac {d^{2}}{dt^{2}}}{\frac {1}{t}}$

$\int _{0}^{\infty }{\frac {d^{2}}{dt^{2}}}e^{-tx}dx={\frac {d^{2}}{dt^{2}}}{\frac {1}{t}}$

$\int _{0}^{\infty }{\frac {d}{dt}}\left(-xe^{-tx}\right)dx={\frac {d}{dt}}\left(-{\frac {1}{t^{2}}}\right)$

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

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

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