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