# 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 $\displaystyle\int_{0}^{\infty}{e^{-ax}dx}=\frac{1}{a}$ …

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

[Note] : Feynman applied “The Fundamental Theorem of Calculus”: Differentiation (aka Derivative) is inverse of Integration (aka Anti-Derivative) , vice versa. $\displaystyle\frac{d}{da}\displaystyle\int_{0}^{\infty}e^{-ax}dx= \int_{0}^{\infty}\frac{d}{da}(e^{-ax})dx=\int_{0}^{\infty} -xe^{-ax}dx$

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

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]

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## 2 thoughts on “Differentiating under integral”

1. tomcircle

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