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I consider the following function on $\mathbb{R}$ :

$$ f(z)=\int\limits_0^{+\infty} dx\frac{1+x}{\sqrt{x}}\frac{1}{x\, e^x +z}$$

I would like to know its singular expansion at $z=0$.

Doing a series expansion of the denominator does not help me much, so I don't really know where to start...

Alexandre Krajenbrink
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  • What is you variable of integration? – MrYouMath Oct 14 '16 at 13:58
  • It is $x$, sorry for the misprint, I just corrected it ! – Alexandre Krajenbrink Oct 14 '16 at 13:59
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    Have you checked convergence? Why don't you simply determine the taylor series of your integrand and then integrate all terms. The variable $z$ is just a parameter in the expresion. I don't think that there is any simple expression for this series. – MrYouMath Oct 14 '16 at 14:03
  • As $z\rightarrow +\infty$, there is a simple asymptotics for this function - $$f(z) \sim \frac{2}{3} \frac{\ln(z)^{3/2}}{z}+\mathcal{O}(\frac{\ln(z)^{1/2}}{z})$$ (the proof is similar to the proof of the asymptotics of the polylogarithm function, see Wood 1992 ) So I thought I could also have an expansion around 0 ...

    The integral will be divergent at z=0 (unless you use the analytic continuation of the Gamma $\Gamma$ function...)

     – Alexandre Krajenbrink Oct 14 '16 at 14:08

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I don't know if this is accurate enough for your purposes, but just to make some progress, we can proceed as follows. First of all, $\int_1^{\infty}\ldots$ is of course holomorphic near $z=0$, so has a Taylor expansion about $z=0$. Next, we can simplify and consider $$ I(z)= \int_0^1\frac{1+x}{\sqrt{x}}\frac{dx}{x+z} ; $$ the difference between $I$ and the actual integral (over $(0,1)$) stays bounded as $z\to 0+$.

Now $I$ is easy to analyze. We can substitute $x=t^2$ to obtain that $$ I(z) = 2\int_0^1 \frac{t^2+1}{t^2+z}\, dt = 2 + 2(1-z)\int_0^1 \frac{dt}{t^2+z}=\frac{\pi}{\sqrt{z}}+O(1) , $$ and, as discussed, this is also the asymptotic behavior of $f(z)$ as $z\to 0+$.