Find the Limits of Integration for the Gamma Function

eestep
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Homework Statement


Gamma function is defined for all x>0 by rule
\Gamma(x)=\int0\inftytx-1e-tdt
Find a simple expression for \Gamma(n) for positive integers n. Answer is \Gamma(n)=(n-1)!


Homework Equations





The Attempt at a Solution


\int0\inftytn-1e-tdt=-tn-1e-t-\int(n-1)tn-2(-e-t)dt=-tn-1e-t+(n-1)\inttn-2e-tdt
u=tn-1 du=(n-1)tn-2dt
dv=e-tdt v=\inte-tdt=-e-t
 
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Use mathematical induction together with integration by parts.
 
I appreciate the advice!
 
eestep said:

Homework Statement


Gamma function is defined for all x>0 by rule
\Gamma(x)=\int_0^\infty\, t^{x-1}\,e^{-t}\,dt
Find a simple expression for \Gamma(n) for positive integers n. Answer is \Gamma(n)=(n-1)!

Homework Equations



The Attempt at a Solution



\Gamma(n)=\int_0^\infty\, t^{n-1}\,e^{-t}\,dt=-t^{n-1}\,e^{-t}-\int(n-1)t^{n-2}(-e^{-t})dt=-t^{n-1}e^{-t}\ +\ (n-1)\int t^{n-2}e^{-t}dt<br />

You are missing your limits of integration after doing integration by parts.

Click on the expression at the right to see the LaTeX code that produced it: \left[a^{-x}\right]_{\sqrt{2}}^{\infty}  .
 
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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