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The relevant part of the integral definition of $Y_0$ is

$$-\frac{2}{\pi}\int_0^{\infty } e^ {-x \sinh t} \, dt$$

which should be asymptotic to

$$\frac{2}{\pi}\big( \ln\frac x 2 + \gamma \big)$$

Where $\gamma$ is the Euler-Mascheroni Constant. How can I evaluate that integral for small x?

It would be sufficient to find the logarithmic term, but I would also be interested in where the Euler-Mascheroni Constant comes from.

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  • $\begingroup$ Use taylor expansion or pade expansion in x? $\endgroup$
    – MrYouMath
    Nov 2, 2016 at 21:06
  • $\begingroup$ But I can't integrate sinh from 0 to infinity $\endgroup$ Nov 2, 2016 at 21:12
  • $\begingroup$ Ah you are right totally ignored that :D. $\endgroup$
    – MrYouMath
    Nov 2, 2016 at 21:15
  • $\begingroup$ @Fabian yes, thanks $\endgroup$ Nov 2, 2016 at 21:29

3 Answers 3

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The intuition is that for small $x$ only the contribution for "large" $t$ is important. Thus, we expect that $\sinh t \sim \tfrac12 e^{t}$. So we approximate $$Y_0(x) \sim - \frac2{\pi} \int_0^\infty\!dt\,e^{-x e^{t}/2}= - \frac{2}{\pi}\int_{x/2}^\infty \!dz \frac{e^{-z}}{z}. \tag{1}$$

Indeed, we have that \begin{align*}\left|Y_0(x) +\frac2{\pi} \int_0^\infty\!dt\,e^{-x e^{t}/2}\right| & \leq \frac{2}{\pi} \int_0^\infty \!dt\,\left| e^{-x e^t/2} - e^{-x \sinh t} \right| \\ & = \frac{2}{\pi} \int_0^\infty \!dt\, e^{-x e^t/2} \underbrace{(1- e^{-x e^{-t}/2})}_{\leq x e^{-t}/2}=O(x) \end{align*} such that $(1)$ is correct (up and with the constant term).

Starting with $(1)$, we can proceed as follows: $$Y_0(x) \sim I=- \frac{2}{\pi}\int_{x/2}^\infty \!dz \frac{e^{-z}}{z}$$ We calculate $$\frac{dI}{dx} = \frac{e^{-x/2}}{\pi (x/2)} = \frac{1}{\pi(x/2)} + O(1)$$ and thus $$I = \frac{2}{\pi} \ln(x/2) + C $$ with $C$ a constant.

The constant is given by \begin{align*} C= \lim_{x\to0^+}\left[I - \frac{2}{\pi} \ln(x/2) \right]&=\frac{2}{\pi} \lim_{x\to0^+}\int_{x/2}^1 \frac{1-e^{-z}}z - \frac{2}\pi \int_1^\infty \frac{e^{-z}}z \\ & = \frac{2}{\pi} \int_{0}^1 \frac{1-e^{-z}}z - \frac{2}\pi \int_1^\infty \frac{e^{-z}}z .\end{align*} I am not sure there is an easy way to show that $C = 2 \gamma/\pi$.

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    $\begingroup$ Euler-Mascheroni Constant is defined as the difference between an integral and a summation. Changing the variables & expanding one of the integrals might do the trick. $\endgroup$ Nov 2, 2016 at 22:49
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    $\begingroup$ since the last integral equal $\text{Ei}(-1)$ which has no closed form something strange is going on here, since the integral $I=2 \text{Ei}(x/2)/\pi$ indeed gives the correct asymptotics wolframalpha.com/input/… $\endgroup$
    – tired
    Nov 3, 2016 at 8:09
  • $\begingroup$ Constant is fixed, see my answer below $\endgroup$
    – tired
    Nov 3, 2016 at 13:32
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The constant might be fixed the following way:

Denote the integral in question is $D=\frac{\pi C}{2}=\color{blue}{J_1}+\color{red}{J_2}$ with

$$ \color{blue}{J_1}=\color{blue}{\int_0^1\frac{1-e^{-z}}{z}\quad},\quad\color{red}{J_2}=\color{red}{-\int_1^{\infty}\frac{e^{-z}}{z}\quad} $$

Let us start with $\color{blue}{J_1}$ and integrate by parts:

$$ \color{blue}{J_1}=\color{blue}{\lim_{\epsilon\rightarrow0}\left(\log{\epsilon}-\int_{\epsilon}^1\frac{e^{-z}}{z}\right)}=\color{blue}{\lim_{\epsilon\rightarrow0}\left(\log{\epsilon}-\log{\epsilon}-\int_{0}^1\log(z)e^{-z}\right)}=\color{blue}{-\int_{0}^1\log(z)e^{-z}} $$

Integrating now $\color{red}{J_2}$ by parts yields

$$ \color{red}{J_2}=\color{red}{-\int_{1}^{\infty}\log(z)e^{-z}} $$

therefore

$$ -D=-(\color{blue}{J_1}+\color{red}{J_2})=\color{blue}{\int_{0}^1\log(z)e^{-z}}+\color{red}{\int_{1}^{\infty}\log(z)e^{-z}}={\int_{0}^{\infty}\log(z)e^{-z}} $$

or

$$ D=\gamma $$

which implies

$$ C=\frac{2 \gamma}{\pi} $$

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    $\begingroup$ Thanks. I just couldn't find a way to proof it yesterday... $\endgroup$
    – Fabian
    Nov 3, 2016 at 17:08
  • $\begingroup$ @Fabian i would say nice teamwork! $\endgroup$
    – tired
    Nov 3, 2016 at 18:06
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Similar to the approach the user Maxim suggested in the comments here, we can use the definition $$Y_{0}(x) = \lim_{\alpha \to 0} \frac{J_{\alpha}(x) \cos(\alpha \pi)-J_{-\alpha}(x)}{\sin(\alpha \pi)}$$

and the fact that $$J_{\alpha}(x) \sim \frac{1}{\Gamma(\alpha +1)} \left(\frac{x}{2}\right)^{\alpha} \quad \text{as} \ x \to 0^{+}$$ for $\alpha>-1$.

Applying L'Hôpital's rule, we get $$\begin{align} Y_{0}(x) &\sim \lim_{\alpha \to 0} \frac{\frac{1}{\Gamma(\alpha +1)} \left(\frac{x}{2}\right)^{\alpha} \cos(\alpha \pi)-\frac{1}{\Gamma(1-\alpha )} \left(\frac{x}{2}\right)^{-\alpha} }{\sin(\alpha \pi)} \\ &= \small \lim_{\alpha \to 0}\frac{-\frac{\Gamma'(\alpha+1)}{\Gamma^{2}(\alpha +1)}\left(\frac{x}{2}\right)^{\alpha} \cos(\alpha \pi) + \frac{\cos(\alpha \pi)}{\Gamma(\alpha+1) }\left(\frac{x}{2}\right)^{\alpha} \ln \left(\frac{x}{2}\right) - \frac{\pi\sin(\alpha \pi)}{\Gamma(\alpha +1) }\left(\frac{x}{2}\right)^{\alpha} -\frac{\Gamma'(1-\alpha)}{\Gamma^{2}(1-\alpha)} \left(\frac{x}{2} \right)^{-\alpha} + \frac{\left(\frac{x}{2} \right)^{-\alpha}}{\Gamma(1-\alpha)} \ln \left(\frac{x}{2} \right) }{\pi \cos(\alpha \pi)} \\ &= \frac{-\Gamma'(1) +\ln \left(\frac{x}{2} \right)-0-\Gamma'(1)+ \ln \left(\frac{x}{2} \right)}{\pi} \\ &= \frac{-2 \Gamma'(1) + 2\ln \left(\frac{x}{2} \right) }{\pi} \\ &= \frac{2}{\pi} \left(\gamma+ \ln \left(\frac{x}{2} \right)\right). \end{align}$$

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