Calculating $\int_{C^+(2,2)} \frac {e^\sqrt z} {(z-2)^2}dz$ and $\int_{0}^{\infty} \frac 1 {1+x \sqrt x}dx$ I want to calculate $$\int_{C^+(2,2)} \frac {e^\sqrt z} {(z-2)^2}dz\quad\mbox{and}\quad\int_{0}^{\infty} \frac 1 {1+x \sqrt x}dx$$ using complex integration. In the first part $\sqrt z$ denotes the principal value. Can somebody guide me how to use the Residue theorem here.
 A: The second integral may be transformed into a standard integral used in residue theory by the substitution $u=x^{3/2}$ to get
$$\int_0^{\infty} \frac{dx}{1+x \sqrt{x}} = \frac{2}{3} \int_0^{\infty} du \: \frac{u^{-1/3}}{1+u}$$
The integral on the RHS may be evaluated by considering
$$\oint_C dz \: \frac{z^{-1/3}}{1+z}$$
over a keyhole contour about the positive real axis.  The integral over the large and small circular arcs vanish in the limits of the radii of those arcs going to infinity and zero, respectively.  The contour integral, which is equal to $i 2 \pi$ times the residue at the pole $z=-1$, is then equal to the contributions up and back along the positive real axis.  The result is
$$(1-e^{-i 2 \pi/3}) \int_0^{\infty} du \: \frac{u^{-1/3}}{1+u} = i 2 \pi e^{-i \pi/3}$$
I leave it to the reader to work the algebra.  The result is
$$\int_0^{\infty} du \: \frac{u^{-1/3}}{1+u} =\frac{2 \pi}{ \sqrt{3}}$$
and therefore
$$\int_0^{\infty} \frac{dx}{1+x \sqrt{x}} = \frac{4 \pi}{3 \sqrt{3}}$$
A: I'm guessing that $C^+(2,2)$ denotes the positively-oriented circle of radius $2$ centred at $2$.  Use the generalized Cauchy integral formula
$$ f^{(n)}(a) = \frac{n!}{2\pi i}\oint_\Gamma \frac{f(z)}{(z-a)^{n+1}}$$
