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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

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

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    exactly, both integrands are to be over (-Inf,Inf).2012-12-30

2 Answers 2

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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} $

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    I edited my question, giving my resolution. Un$f$ortunately I could not give an answer to my own question, because o$f$ having too less reputation.2012-12-30
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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