Gamma function calculation

1. Oct 14, 2011

matematikuvol

$$\Gamma(x)=\int^{\infty}_0t^{x-1}e^{-t}dt$$

$$\Gamma(\frac{1}{2})=\int^{\infty}_0\frac{e^{-t}}{\sqrt{t}}dt=$$

take $$t=x^2$$

$$dt=2xdx$$

$$x=\sqrt{t}$$

$$=\int^{\infty}_0\frac{e^{-x^2}}{x}2xdx$$

Why here we can here reducing integrand by $$x$$?

2. Oct 14, 2011

phyzguy

What is your question? You've done the substitution correctly, so you see how the x's cancel. What is the problem?

3. Oct 14, 2011

matematikuvol

Lower limit is $$0$$. Why I may cancel x's?

4. Oct 14, 2011

dextercioby

Because the functions are defined on intervals not containing the point 0, that's why you can have the x in the denominator and simplify it through.