What's the value of this integral $\int_\mathbb{C}q^{|z|}dz$?

Question

Mathematica cannot take it.

$$\int_0^{\infty } \left(\int_0^{\infty } q^{\sqrt{x^2+y^2}} \, dx\right) \, dy$$

In general, I am interested in the following problem:

$$\int_\mathbb{C}q^{|z|}dz$$

In the both cases one can assume $0<q<1$.

Answer 1

Using polar form should do it

$$ \int_{0}^{\infty}{\int_{0}^{\infty}{q^{\sqrt{x^{2}+y^{2}}}dx}dy}=\int_{0}^{\frac{\pi}{2}}{\int_{0}^{\infty}{rq^{r}dr}d\theta} $$

can You solve this now?

note: remember that You are only integrating over the first quadrant, so it should be $\int_{0}^{\frac{\pi}{2}}$ instead of $\int_{0}^{2\pi}$

Answer 2

Let

$$z=x+iy$$

Then our integral is just

$$\int\limits_{-\infty}^\infty\int\limits_{-\infty}^\infty q^{|x+iy|}\text{d}x\text{d}y$$

which is just

$$\int\limits_{-\infty}^\infty\int\limits_{-\infty}^\infty q^{\sqrt{x^2+y^2}}\text{d}x\text{d}y$$

Switch to polar coordinates:

$$\int\limits_{0}^{2\pi}\int\limits_{0}^\infty rq^{r}\text{d}r\text{d}\theta$$

From your restrictions on $q$, we get

$$\frac{2\pi}{\ln^2(q)}$$

Answer 3

Use the substitution $x=r\cos\theta,y=r\sin\theta,$ which gives

$$\int_{-\infty}^{\infty } \left(\int_{-\infty}^{\infty } q^{\sqrt{x^2+y^2}} \, dx\right) \, dy=\int_0^{2\pi } \left(\int_0^{\infty } q^{r} r\, dr\right) \, d\theta=\frac{2\pi}{\ln q}\int_0^\infty (q^r)'rdr=\frac{2\pi}{\ln q}([q^rr]_0^\infty-\int_0^\infty q^rdr)=-\frac{2\pi}{(\ln q)^2}[q^r]^\infty_0=\frac{2\pi}{(\ln q)^2}.$$