Differentiating under integral

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

Proof:
∀ a>0
Integrate by parts:

$latex displaystyleint_{0}^{infty}{e^{-ax}dx}=-frac{1}{a}e^{-ax}Bigr|_{0}^{infty}=frac{1}{a}$

∀ a>0
$latex displaystyleint_{0}^{infty}{e^{-ax}dx}=frac{1}{a}$ …[1]

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

$latex displaystylefrac{d}{da}displaystyleint_{0}^{infty}e^{-ax}dx= int_{0}^{infty}frac{d}{da}(e^{-ax})dx=int_{0}^{infty} -xe^{-ax}dx$

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

Continue to differentiate with respect to ‘a’:
$latex -2a^{-3} =int_{0}^{infty}-x^{2}e^{-ax}dx$
$latex 2a^{-3} =int_{0}^{infty}x^{2}e^{-ax}dx$
$latex frac{d}{da} text{ both sides}$
$latex 2.3a^{-4} =int_{0}^{infty}x^{3}e^{-ax}dx$
…
…
$latex 2.3.4dots n.a^{-(n+1)} =int_{0}^{infty}x^{n}e^{-ax}dx$
Set a = 1
$latex boxed{n!=int_{0}^{infty}x^{n}e^{-x}dx}$ [QED]

Another Example using “Feynman Integration”:

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

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

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

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

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

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