Integration using parametric derivatives

Integration by parametric derivatives is a method of integrating certain functions.

Suppose you wanted to find the integral:

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

We may solve this by starting with the integral:

$\int _{0}^{\infty }e^{-tx}dx=$

$\left[{\frac {e^{-tx}}{-t}}\right]_{0}^{\infty }=$

$\left[\lim _{x\to \infty }{\frac {e^{-tx}}{-t}}\right]-\left[{\frac {e^{-t0}}{-t}}\right]=$

$\left[0\right]-\left[{\frac {1}{-t}}\right]={\frac {1}{t}}$

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:

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

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

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

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

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:

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