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I am trying to solve the following integral;

$\int_0^\infty \frac{1-\cos(x)}{x \left(\left(\frac{x}{\Lambda}\right)^2 +1\right)^{\frac{1}{2}}}\,dx = I(\Lambda), \Lambda > 0$

My approach thus far has been to split the integral at $\Lambda$ such that the fraction $\frac{x}{\Lambda}$ is either greater than or less than 1. By then considering limiting cases of $\Lambda$, for $\Lambda \gg 1$ I obtain a logarithmic dependence $I(\Lambda) \sim log(\Lambda)$ (the dominant integral looks like the Cin(x) function) and for $\Lambda \ll 1$ there is a linear dependence $I(\Lambda) \sim \frac{\pi \Lambda}{2}$. This agrees with what I have seen by doing the integral numerically.

After reading through some papers I stumble upon the integral and an exact analytic solution having the form;

$I(\Lambda) = \frac{\pi \Lambda}{2} {}_{1}F_{2}\left(1/2;1,3/2;\Lambda^2 \right) - (\Lambda^2){}_{2}F_{3}\left( 1,1;3/2,3/2,2;\Lambda^2\right)$

Where ${}_{p}F_{q}$ are generalised hypergoemetric functions. My problem is that I am unable to show why this is the fully analytic solution.

My only thoughts so far is along the lines of a Mellin-Barnes representation for the sqaure root term, splitting the integral again as described in the answer in the link. Then perhaps after some integration one is left with the Barnes integral representation of the hypergeometric functions.

Is this on the right lines of solving the problem? I am not familiar with Mellin transforms and Barnes integral representation, so any enlightenment on the issue, or any other possible direction of solving the integral would be a great help.

Thanks

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Let's clean a bit the integral with

$$\frac{x}{\Lambda} = y$$

Which gives you

$$\int_0^{+\infty} \frac{1 - \cos(\Lambda x)}{y\sqrt{y^2 + 1}}\ dy$$

At this point we can use Taylor Series for the cosine:

$$\cos(\Lambda y) = \sum_{k = 0}^{+\infty} \frac{(-1)^k (\Lambda y)^{2k}}{(2k)!}$$

And we can also notice that

$$1 - \cos(\Lambda y) \equiv \sum_{k = 1}^{+\infty} \frac{(-1)^k (\Lambda y)^{2k}}{(2k)!}$$

Hence the integral becomes

$$\sum_{k = 1}^{+\infty} \frac{(-1)^k (\Lambda)^{2k}}{(2k)!} \int_0^{+\infty} \frac{y^{2k}}{y\sqrt{y^2 + 1}}\ dy = \sum_{k = 1}^{+\infty} \frac{(-1)^k (\Lambda)^{2k}}{(2k)!} \int_0^{+\infty} \frac{y^{2k-1}}{\sqrt{y^2 + 1}}\ dy $$

The latter integral is very straightforward if you have some basics of Gamma function methods, indeed:

$$\int_0^{+\infty} \frac{y^{2k-1}}{\sqrt{y^2 + 1}}\ dy = \frac{\Gamma\left(\frac{1}{2} - k\right)\Gamma(k)}{2\sqrt{\pi}}$$

Putting all together we get:

$$\frac{1}{2\sqrt{\pi}} \sum_{k = 1}^{+\infty} \frac{(-1)^k (\Lambda)^{2k}}{(2k)!}\Gamma\left(\frac{1}{2} - k\right)\Gamma(k)$$

That series can be summed, and again through some knowledge of Special functions (and Hypergeometric functions) we arrive to:

$$I(\Lambda) = \boxed{\sqrt{\pi } \Lambda^2 \, _2F_3\left(1,1;\frac{3}{2},\frac{3}{2},2;\frac{\Lambda^2}{4}\right)}$$

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    $\begingroup$ The integral $\int_0^{+\infty} \frac{y^{2k-1}}{\sqrt{y^2 + 1}}\ dy$ is badly divergent for $k>1/2$, though... $\endgroup$ – Pierpaolo Vivo Mar 6 '18 at 12:18
  • $\begingroup$ @PierpaoloVivo Ah that is indeed a good observation, lol. Taylor Series seemed to be a good compromise, but probably I have to rethink about... $\endgroup$ – Von Neumann Mar 6 '18 at 12:24
  • $\begingroup$ @VonNeumann thanks for the help, could you please point me in the direction of some texts to understand the gamma function methods mentioned in your attempted solution? $\endgroup$ – Tim Mar 18 '18 at 15:50
  • $\begingroup$ @Tim Sure. Gamma function is related to many other special function, hence books on special functions are better. That being said, and you can use Wong's Special Functions and Lebedev one. Also you can find great notes online for free. For very detailed identities and relations, then NIST handbook od Special functions is the top, together with Abramowitz-Stegun Classics book of special functions! $\endgroup$ – Von Neumann Mar 18 '18 at 15:53

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