Gamma function calculation

[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 $$x$$?

 

[phyzguy]

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

 

[matematikuvol]

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

 

[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.