Integral Simplification

Question

I was hoping to get some help....

I have a complex integral expression: $\frac{\int_0^\infty t\left( A(0)\alpha \left( b+1 \right) {e}^{-\alpha t \left( b+1\right)} + A(0) \frac{b}{g - 1- b } \left( \alpha \left( b+1\right) {e}^{-\alpha t \left( b+1\right)} - \alpha g {e}^{-\alpha t g}\right) -\left( A0-1\right) \alpha g {e}^{-\alpha t g} \right) \mathrm{d}t} {\int_0^\infty t\left(\alpha {e}^{-\alpha t}\right) \mathrm{d}t}$

I want to eliminate $\alpha$ from the expression. My calculus skills are rudimentary... is there any way I can do this? I tried U-substitution and couldn't find a way to eliminate $\alpha$.

Thank you

Mark

Answer 1

OK thanks,

Figured in out

$\frac{A(0)g (g - 1- b)+A(0)bg-A(0)b(1+b)+(1-A(0))(1+b) (g - 1- b)}{(1+b)g (g - 1- b)}$

Answer 2

You just need to know one integral: $$ \int_{0}^{\infty}t{\rm e}^{-\mu t}\,{\rm d}t = {1 \over \mu^{2}}\,\qquad \mu > 0 $$