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

Example One: Exponential Integral

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