Why is the gamma function equal to (n-1) ?

michonamona
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Homework Statement



Why is the equality below true?

\Gamma(n) = (n-1)!

Where \Gamma(n) = \int^{\infty}_{0} x^{n-1} e^{-x}dx

Homework Equations


The Attempt at a Solution



I've read the article on wikipedia but I cannot understand it. Is there any special properties in calculus that I must know in order to comprehend this?

Thank you
M
 
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It is true only if n is integer

Solve:

\int_0^\infty t^{4-1}e^{-t} dt

Compare if it's equal to
(4-1)!

Gamma function is really beautiful beacuse it extends the concept of factorial out of integer numbers.
I've read the article on wikipedia but I cannot understand it. Is there any special properties in calculus that I must know in order to comprehend this?

Yep, being able to integrate. :()
 
You can use integration by parts to show that

\Gamma(n) = (n-1)\Gamma(n-1)

If n is an integer, you can use this to prove by induction that

\Gamma(n) = (n-1)!
 
I went to a talk by John Chapman on this and he said said that the Gamma function is related to "Runge phenomenon".
 
Prove $$\int\limits_0^{\sqrt2/4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx = \frac{\pi^2}{8}.$$ Let $$I = \int\limits_0^{\sqrt 2 / 4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx. \tag{1}$$ The representation integral of ##\arcsin## is $$\arcsin u = \int\limits_{0}^{1} \frac{\mathrm dt}{\sqrt{1-t^2}}, \qquad 0 \leqslant u \leqslant 1.$$ Plugging identity above into ##(1)## with ##u...
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