# Differentiating under integral

Prove: (Euler Gamma Γ Function)
$\displaystyle n! = \int_{0}^{\infty}{x^{n}.e^{-x}dx}$

Proof:
∀ a>0
Integrate by parts:

$\displaystyle\int_{0}^{\infty}{e^{-ax}dx}=-\frac{1}{a}e^{-ax}\Bigr|_{0}^{\infty}=\frac{1}{a}$

∀ a>0
$\displaystyle\int_{0}^{\infty}{e^{-ax}dx}=\frac{1}{a}$ …[1]

Feynman trick: differentiating under integral => d/da left side of [1]

$\displaystyle\frac{d}{da}\displaystyle\int_{0}^{\infty}e^{-ax}dx= \int_{0}^{\infty}\frac{d}{da}(e^{-ax})dx=\int_{0}^{\infty} -xe^{-ax}dx$

Differentiate the right side of [1]:
$\displaystyle\frac{d}{da}(\frac{1}{a}) = -\frac{1}{a^2}$
=>
$a^{-2}=\int_{0}^{\infty}xe^{-ax}dx$

Continue to differentiate with respect to ‘a’:
$-2a^{-3} =\int_{0}^{\infty}-x^{2}e^{-ax}dx$
$2a^{-3} =\int_{0}^{\infty}x^{2}e^{-ax}dx$
$\frac{d}{da} \text{ both sides}$
$2.3a^{-4} =\int_{0}^{\infty}x^{3}e^{-ax}dx$

$2.3.4\dots n.a^{-(n+1)} =\int_{0}^{\infty}x^{n}e^{-ax}dx$
Set a = 1
$\boxed{n!=\int_{0}^{\infty}x^{n}e^{-x}dx}$ [QED]

Another Example using “Feynman Integration”:

$\displaystyle \text{Evaluate }\int_{0}^{1}\frac{x^{2}-1}{ln x} dx$

$\displaystyle \text{Let I(b)} = \int_{0}^{1}\frac{x^{b}-1}{ln x} dx$ ; for b > -1

$\displaystyle \text{I'(b)} = \frac{d}{db}\int_{0}^{1}\frac{x^{b}-1}{ln x} dx = \int_{0}^{1}\frac{d}{db}(\frac{x^{b}-1}{ln x}) dx$

$x^{b} = e^{ln x^{b}} = e^{b.ln x}$

$\frac{d}{db}(x^{b}) = \frac{d}{db}e^{b.ln x}=e^{b.ln x}.{ln x}= e^{ln x^{b}}.{ln x}=x^{b}.{ln x}$

$\text{I'(b)}=\int_{0}^{1} x^{b} dx=\frac{x^{b+1}}{b+1}\Bigr|_{0}^{1} = \frac{1}{b+1}$
=>
$\text {I(b)} = ln (b+1) + C$

Let b=0
I(0) = 0= ln (1) + C = 0+C => C=0

$\boxed{I(b)=ln(b+1)}$

Let b= 2
$\displaystyle\int_{0}^{1}\frac{x^{2}-1}{ln x} dx = I(2) = ln (3)$ [QED]