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I stumbled on an interesting integral doing some physics exercise which did not require its closed form (if it has any). It has, however, sparked my interest and I tried my best to find it, but I couldn't. The integral is :

$$\int_0^R \frac{\text{d}x}{\sqrt{\ln(1+x)}}$$ with $$R>0$$

It looks like $$\int_0^R \frac{\text{d}x}{(x+1)\sqrt{\ln(1+x)}}$$ which is easy to integrate, but it seems like the first one is a lot harder. Wolfram Alpha does not display any results in the general case (for any positive real $R$), even if it does find some results for characteristic values like $R$=1. This makes me think there is no easy closed form for this integral. I am, however, open to any suggestions on how to tackle this more efficiently, or any solution, even if it is probably beyond my maths level.

Edit : This wasn't so hard indeed, I should have thought about this. I'm still puzzled about the complex substitution necessary to find tired's result (with a missing factor of $-i$ apparently), so I'll look for some confirmation before I accept anything.

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  • $\begingroup$ this integral is equivalent to an special function called "error integral". Btw: in which physical context does this integral appear? $\endgroup$
    – tired
    Oct 9, 2015 at 17:06
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    $\begingroup$ This integral appears in the context of an "exploding" cylinder of electrons. We start with a collimated, infinite cylinder of electrons of radius r, then it expands under the effects of the Coulomb repulsion. Neglecting the magnetic field and admitting there is no movement in the axis direction (meaning the electrons are pushed away from the axis radially), the radius expands like r+ξ(r,t), and we can find that the radius doubles for T=1/(wp)*integral from 0 to 1 of dx/sqrt(ln(1+x)), where wp is the Langmuir frequency.(of course, x has no dimension here) $\endgroup$
    – Evariste
    Oct 9, 2015 at 17:13
  • $\begingroup$ this is interesting! never stumbeled over such an integral during my research! $\endgroup$
    – tired
    Oct 9, 2015 at 17:20

1 Answer 1

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Ok, this is not so difficult:

Setting $x+1=y$ our integral in question is:

$$ I=\int_1^{R+1}\frac{1}{\sqrt{\log(y)}}dy $$

next, using $y=e^q$ we obtain:

$$ I=\int_0^{\log(R+1)}\frac{e^q}{\sqrt{q}}dq $$

then, setting $q=p^2$ we get

$$ I=2\int_{0}^{\sqrt{\log(R+1)}}e^{p^2}dp=-i\sqrt{\pi}\text{Erf}(i\sqrt{\log(R+1)}) $$ if I’m not mistaken. $\text{Erf}(z)$ denotes the Error function. Note that it is legitimate to substitute $p\rightarrow ip$ because our integrand happens to be an entire function.

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  • $\begingroup$ Okay I'm gonna do some checking, but I think a "-i" factor is missing. $\endgroup$
    – Evariste
    Oct 9, 2015 at 17:57
  • $\begingroup$ Okay, substituting u=ip assuming it is legitimate (I am not sure about it), the result I find is $I=-i\sqrt(\pi)Erf(i\sqrt{ln(R+1)})$. $\endgroup$
    – Evariste
    Oct 9, 2015 at 18:05
  • $\begingroup$ @Evariste yes u are right. Sorry, i was a little bit fast when i wrote this answer. i corrected it and gave some justification. $\endgroup$
    – tired
    Oct 9, 2015 at 18:43
  • $\begingroup$ It is a theorem of Liouville that $\int e^{x^2}$ cannot be given in terms of the "traditional" functions. $\endgroup$
    – DanielWainfleet
    Oct 9, 2015 at 19:16

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