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 $\phi_{11}(k)$:

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


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

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

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

  • $\begingroup$ It would be easier to read your question if you used LaTeX formatting. $\endgroup$
    – Eckhard
    Dec 30, 2012 at 18:48
  • $\begingroup$ it's my first question and I did not know how to post it properly. Having a look at other posts, I figured out how to post it. Thanks $\endgroup$
    – fpe
    Dec 30, 2012 at 18:51
  • $\begingroup$ You're welcome and welcome to math.SE. You can always edit your question to make it more reader-friendly. $\endgroup$
    – Eckhard
    Dec 30, 2012 at 18:52
  • $\begingroup$ I did as well but strangely it was rejected by "Community" $\endgroup$
    – user50407
    Dec 30, 2012 at 19:08
  • $\begingroup$ thanks everybody, I was just writing my own with LaTeX, but the version from @adamW is already right. $\endgroup$
    – fpe
    Dec 30, 2012 at 19:11

2 Answers 2


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): $$ \begin{eqnarray} F_{11}(k_1)&=&\frac{1}{2}\alpha\epsilon^{2/3}L^{5/3}\left(1+k_1^2\right)^{-5/6}\int_0^\infty \frac{r^3 \, dr}{\left(1+r^2\right)^{17/6}} \\ &=& \frac{9}{55}\alpha\epsilon^{2/3}L^{5/3}\left(1+k_1^2\right)^{-5/6}. \end{eqnarray} $$

  • $\begingroup$ I edited my question, giving my resolution. Unfortunately I could not give an answer to my own question, because of having too less reputation. $\endgroup$
    – fpe
    Dec 30, 2012 at 23:15

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


You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .