Calculate a double integral I would like to ask a pretty easy question (at least I believe so). I know that:

$$\phi_{11}(k) = \frac{E(k)}{4\pi k^4}(k^2 - k_1^2)$$
  $$E(k) =  \alpha \epsilon^{\frac{2}{3}}L^{\frac{5}{3}}\frac{k^4}{(1 + k^2)^{\frac{17}{6}}}$$

therefore, substituting the expression of $E(k)$ in $\phi_{11}(k)$:

$$\phi_{11}(k) = \frac{\alpha\epsilon^{\frac{2}{3}}L^{\frac{5}{3}}}{4\pi}\frac{k^2 - k_1^2}{(1 + k^2)^{\frac{17}{6}}}.$$

Furthermore 

$$k = \sqrt{k_1^2 + k_2^2 + k_3^2}$$

hence the above expression becomes

$$\phi_{11}(k_1,k_2,k_3) = \frac{\alpha\epsilon^{\frac{2}{3}}L^{\frac{5}{3}}}{4\pi}\frac{k_2^2 + k_3^2 }{(1 + k_1^2 + k_2^2 + k_3^2)^{\frac{17}{6}}}.$$

The question is how to manually compute the function

$$F_{11}(k_1) = \int\limits_{-\infty}^\infty\int\limits_{-\infty}^\infty \phi_{11}\, \mathrm dk_2 \mathrm dk_3$$

$F_{11}$ is therefore the double integral of $\phi_{11}$ over unlimited range.
Of course, the first step is to write

$$F_{11}(k_1) = 2 \cdot 2 \cdot \int\limits_0^\infty\int\limits_0^\infty \phi_{11}\, \mathrm dk_2\mathrm dk_3$$

 A: The simplest way to approach this is to note that the integrand has radial symmetry: it depends only on $r^2\equiv k_2^2 + k_3^2$.  (In other words, use $k^2=k_1^2+k_{\perp}^2$ in the first place, with the appropriate area element.)  So the general form here is
$$
\int_{-\infty}^\infty \int_{-\infty}^\infty f(x^2+y^2) \, dx \, dy=2\pi\int_{0}^{\infty}rf(r^2) \, dr.
$$
In your case, this results in
$$
F_{11}(k_1)=\frac{1}{2}\alpha\epsilon^{2/3}L^{5/3}\int_{0}^{\infty}\frac{r^3 dr}{\left(1+k_1^2+r^2\right)^{17/6}}
$$
This you can integrate by parts.  Note that even if this integral were intractable, you could rescale using $r\rightarrow r\sqrt{1+k_1^2}$ to find the functional dependence on $k_1$ (which in many cases is all you need):
$$
\begin{eqnarray}
F_{11}(k_1)&=&\frac{1}{2}\alpha\epsilon^{2/3}L^{5/3}\left(1+k_1^2\right)^{-5/6}\int_0^\infty \frac{r^3 \, dr}{\left(1+r^2\right)^{17/6}} \\
&=& \frac{9}{55}\alpha\epsilon^{2/3}L^{5/3}\left(1+k_1^2\right)^{-5/6}.
\end{eqnarray}
$$
A: I solved it in another way, starting from your hint (that I was already
trying to use):
$$ F_{11}(k_1)=\frac{1}{2}\alpha\epsilon^{2/3}L^{5/3}\int_{0}^{\infty}\frac{r^3 dr}{\left(1+k_1^2+r^2\right)^{17/6}} $$
I then called
$$ 1 + k_1^2 = c $$
so that the integral becomes
$$ F_{11}(k_1)=\frac{1}{2}\alpha\epsilon^{2/3}L^{5/3}\int_{0}^{\infty}\frac{r^3 dr}{\left(c+r^2\right)^{17/6}} $$
which results in the final expression
$$ F_{11}(k_{1})=\frac{9}{55}\alpha\varepsilon^{2/3}L^{5/3}\frac{1}{\left(1+k_{1}^{2}\right)^{5/6}} $$
which is exactly the results given within the scientific paper I was reading.
Best regards,
FPE
