Differentiating under integral

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

∀ a>0
Integrate by parts:


∀ 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}

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


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. Pingback: Why 0! = 1 ? What is (1.5)! ? | Math Online Tom Circle

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