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

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

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

Therefor it be can bounded

• What did you try so far? can you calculate the indefinite integrals? – LeeNeverGup Feb 13 '14 at 13:59
• I've understood the first one.The indefinite integral of the second is (e^(1/x + 1))/lnx but I can't work with that. – user128512 Feb 13 '14 at 14:37
• The third I tried in parts but did not succeed-appreciate your help thanks – user128512 Feb 13 '14 at 14:38
• For 3, consider $\int (1+e^{ix})e^{-x}dx$ since $\exp(ix)=\cos(x)+i\sin(x)$ we take the imaginary part of the resulting integral... – Alex Nelson Feb 13 '14 at 14:44
• There is some method that may be helpful for the third. You denote the integral as I, integrating by parts twice, and get some variation of I. Then you have equation in I, that you can solve. In additional, if you want to check your results, you can use Wolfram Alpha:wolframalpha.com/input/… – LeeNeverGup Feb 13 '14 at 14:47

$\displaystyle \frac1{\sqrt{e^x}}=\frac1{(e^x)^\frac12}=e^{-\frac12x}$
The first integrand can be written as $e^{-\frac{x}{2}}$. Then, you can $u$-substitute $u=-\frac{x}{2}$.
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).