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:

:

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:

$frac\{d^2\}\{dt^2\}\; int\_0^\{infty\}\; e^\{-tx\}\; dx\; =\; frac\{d^2\}\{dt^2\}\; frac\{1\}\{t\}$

$int\_0^\{infty\}\; frac\{d^2\}\{dt^2\}\; e^\{-tx\}\; dx\; =\; frac\{d^2\}\{dt^2\}\; frac\{1\}\{t\}$

$int\_0^\{infty\}\; frac\{d\}\{dt\}\; -x\; e^\{-tx\}\; dx\; =\; frac\{d\}\{dt\}\; -frac\{1\}\{t^2\}$

$int\_0^\{infty\}\; x^2\; e^\{-tx\}\; dx\; =\; frac\{2\}\{t^3\}$

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:

$int\_0^\{infty\}\; x^2\; e^\{-3x\}\; dx\; =\; frac\{2\}\{3^3\}\; =\; frac\{2\}\{27\}$