integral of 1/(sqrt(e^x)) from 0 to infinity(Improper integral)

Question

I'm taking Calculus 1 course and I'm having problems with the following integrals(Improper integrals)

1. $\displaystyle\int_0^{\infty} \frac1{\sqrt{e^x}}$ dx
2. $\displaystyle\int_0^1 e^\frac1x$ dx
3. $\displaystyle\int_1^\infty (1+\sin x)\cdot e^{-x}$ dx

Answer for third:

$\displaystyle\int_1^\infty (1+\sin x)\cdot e^{-x}$ dx < $\displaystyle\int_1^\infty (2)\cdot e^{-x}$ dx = 1

Therefor it be can bounded

Thanks for your help-(helpful sites will also be appriciated)

Answer 1

HINT for 1:

$\displaystyle \frac1{\sqrt{e^x}}=\frac1{(e^x)^\frac12}=e^{-\frac12x}$

Answer 2

The first integrand can be written as $e^{-\frac{x}{2}}$. Then, you can $u$-substitute $u=-\frac{x}{2}$.

Are you sure you wrote down the second integrand right?

The third integrand's $\int e^{-x}$ term is straightforward to do. $\int e^{-x} \sin x$ can be done via integration by parts (similar to the way in this link) or a table (See #117 and u-substitute $u=-x$ first, and note $\sin(-x) = - \sin(x)$ to match the form of the integrands).