Integration using parametric derivatives

In calculus, integration by parametric derivatives, also called parametric integration, is a method of Using known Integrals to integrate derived functions. It is often used in Physics, and is similar to integration by substitution.

Examples

One

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.

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

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

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

Two

Starting with the integral $\int _{-\infty }^{\infty }e^{-x^{2}t}dx={\frac {\sqrt {\pi }}{\sqrt {t}}}$ Taking the derivative with respect to t of both sides
${\begin{aligned}&{\frac {d}{dt}}\int _{-\infty }^{\infty }e^{-x^{2}t}dx={\frac {d}{dt}}{\frac {\sqrt {\pi }}{\sqrt {t}}}\\&-\int _{-\infty }^{\infty }x^{2}e^{-x^{2}t}=-{\frac {\sqrt {\pi }}{2}}t^{\frac {3}{2}}\\&\int _{-\infty }^{\infty }x^{2}e^{-x^{2}t}={\frac {\sqrt {\pi }}{2}}t^{\frac {3}{2}}\end{aligned}}$
In general, Taking the n-th derivative with respect to t gives us
$\int _{-\infty }^{\infty }x^{2n}e^{-x^{2}t}={\frac {(2n-1)!!{\sqrt {\pi }}}{2^{n}}}t^{\frac {2n-1}{2}}$

External links

WikiBooks: Parametric_Integration