Integration using parametric derivatives

In calculus, integration by parametric derivatives, also called parametric integration,^[1] is a method which uses known Integrals to integrate derived functions. It is often used in Physics, and is similar to integration by substitution.

By using the Leibniz integral rule with the upper and lower bounds fixed we get that
${\displaystyle {\frac {d}{dt}}\left(\int _{a}^{b}f(x,t)dx\right)=\int _{a}^{b}{\frac {\partial }{\partial t}}f(x,t)dx}${\displaystyle {\frac {d}{dt}}\left(\int _{a}^{b}f(x,t)dx\right)=\int _{a}^{b}{\frac {\partial }{\partial t}}f(x,t)dx}
It is also true for non-finite bounds.

For example, suppose we want to find the integral

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

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

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

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

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

Starting with the integral ${\displaystyle \int _{-\infty }^{\infty }e^{-x^{2}t}dx={\frac {\sqrt {\pi }}{\sqrt {t}}}}${\displaystyle \int _{-\infty }^{\infty }e^{-x^{2}t}dx={\frac {\sqrt {\pi }}{\sqrt {t}}}}, taking the derivative with respect to t on both sides yields
${\displaystyle {\begin{aligned}&{\frac {d}{dt}}\int _{-\infty }^{\infty }e^{-x^{2}t}dx={\frac {d}{dt}}{\frac {\sqrt {\pi }}{\sqrt {t}}}\\&-\int _{-\infty }^{\infty }x^{2}e^{-x^{2}t}=-{\frac {\sqrt {\pi }}{2}}t^{-{\frac {3}{2}}}\\&\int _{-\infty }^{\infty }x^{2}e^{-x^{2}t}={\frac {\sqrt {\pi }}{2}}t^{-{\frac {3}{2}}}\end{aligned}}}${\displaystyle {\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
${\displaystyle \int _{-\infty }^{\infty }x^{2n}e^{-x^{2}t}={\frac {(2n-1)!!{\sqrt {\pi }}}{2^{n}}}t^{-{\frac {2n+1}{2}}}}${\displaystyle \int _{-\infty }^{\infty }x^{2n}e^{-x^{2}t}={\frac {(2n-1)!!{\sqrt {\pi }}}{2^{n}}}t^{-{\frac {2n+1}{2}}}}.

Using the classical ${\displaystyle \int x^{t}dx={\frac {x^{t+1}}{t+1}}}${\displaystyle \int x^{t}dx={\frac {x^{t+1}}{t+1}}} and taking the derivative with respect to t we get
${\displaystyle \int \ln(x)x^{t}={\frac {\ln(x)x^{t+1}}{t+1}}-{\frac {x^{t+1}}{(t+1)^{2}}}}${\displaystyle \int \ln(x)x^{t}={\frac {\ln(x)x^{t+1}}{t+1}}-{\frac {x^{t+1}}{(t+1)^{2}}}}.

The method can also be applied to sums, as exemplified below.
Use the Weierstrass factorization of the sinh function:
${\displaystyle {\frac {\sinh(z)}{z}}=\prod _{n=1}^{\infty }\left({\frac {\pi ^{2}n^{2}+z^{2}}{\pi ^{2}n^{2}}}\right)}${\displaystyle {\frac {\sinh(z)}{z}}=\prod _{n=1}^{\infty }\left({\frac {\pi ^{2}n^{2}+z^{2}}{\pi ^{2}n^{2}}}\right)}.
Take the logarithm:
${\displaystyle \ln(\sinh(z))-\ln(z)=\sum _{n=1}^{\infty }\ln \left({\frac {\pi ^{2}n^{2}+z^{2}}{\pi ^{2}n^{2}}}\right)}${\displaystyle \ln(\sinh(z))-\ln(z)=\sum _{n=1}^{\infty }\ln \left({\frac {\pi ^{2}n^{2}+z^{2}}{\pi ^{2}n^{2}}}\right)}.
Derive with respect to z:
${\displaystyle \coth(z)-{\frac {1}{z}}=\sum _{n=1}^{\infty }{\frac {2z}{z^{2}+\pi ^{2}n^{2}}}}${\displaystyle \coth(z)-{\frac {1}{z}}=\sum _{n=1}^{\infty }{\frac {2z}{z^{2}+\pi ^{2}n^{2}}}}.
Let ${\displaystyle w={\frac {z}{\pi }}}${\displaystyle w={\frac {z}{\pi }}}:
${\displaystyle {\frac {1}{2}}{\frac {\coth(\pi w)}{\pi w}}-{\frac {1}{2}}{\frac {1}{z^{2}}}=\sum _{n=1}^{\infty }{\frac {1}{n^{2}+w^{2}}}}${\displaystyle {\frac {1}{2}}{\frac {\coth(\pi w)}{\pi w}}-{\frac {1}{2}}{\frac {1}{z^{2}}}=\sum _{n=1}^{\infty }{\frac {1}{n^{2}+w^{2}}}}.

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