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

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

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

${\begin{aligned}&{\frac {d^{2}}{dt^{2}}}\int _{0}^{\infty }e^{-tx}\,dx={\frac {d^{2}}{dt^{2}}}{\frac {1}{t}}\\[10pt]&\int _{0}^{\infty }{\frac {d^{2}}{dt^{2}}}e^{-tx}\,dx={\frac {d^{2}}{dt^{2}}}{\frac {1}{t}}\\[10pt]&\int _{0}^{\infty }{\frac {d}{dt}}\left(-xe^{-tx}\right)\,dx={\frac {d}{dt}}\left(-{\frac {1}{t^{2}}}\right)\\[10pt]&\int _{0}^{\infty }x^{2}e^{-tx}\,dx={\frac {2}{t^{3}}}.\end{aligned}}$

Failed to parse (syntax error): {\displaystyle \int_2^\3 x^2 \, dx = frac{e^i/d} This is the same form as the desired integral, where ''t''&nbsp;=&nbsp;3. Substituting that into the above equation gives the value: : <math>\int_0^\infty x^2 e^{-3x} \, dx = \frac{2}{3^3} = \frac{2}{27}.}