Integration using parametric derivatives

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

Suppose we want to find the integral

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

We may solve this by starting with the integral:

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

Now that we know:

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

Suppose we found the second derivative with respect, not to x, but to t:

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

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

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

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

Now notice that this solution takes the same form as the original proposed question. In the original problem, t = 3. Substituting that into our new solution equation:

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