# Find $\lim \limits_{y\rightarrow\infty}\left (\ln^2y\,-2\int_{0}^y\frac{\ln x}{\sqrt{x^2+1}}dx\right)$

I have difficulty with this limit. Where to start?

$$\lim_{y\rightarrow\infty}\left (\ln^2y\,-2\int_{0}^y\frac{\ln x}{\sqrt{x^2+1}}dx\right)$$

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Is the integral from $0$ to $y$? Near zero, the integrand looks like $\log x$, which is not integrable... – Siminore Jun 29 '12 at 14:57
@Siminore, $\log x$ is integrable near $x = 0$. – Sangchul Lee Jun 29 '12 at 14:58
Sorry, in Italy it's 5 pm and I'm tired :-) – Siminore Jun 29 '12 at 15:05
Another solution can be found here – akotronis Jul 1 '12 at 9:49
@Chris, By elementary solution what do you mean? Though demonstrated in somewhat unintuitive fashion, my solution involves only calculus-level manipulations except that it exploited the result $$\zeta(2)=\frac{\pi^2}{6}$$ of the Basel's problem. So I wonder how elementary level you are expecting. – Sangchul Lee Jul 1 '12 at 17:53

By simple integration by parts, we have

$$\int_{0}^{y} \frac{\log x}{\sqrt{1+x^2}} \; dx = \log y \, \sinh^{-1}y - \int_{0}^{y} \frac{\sinh^{-1}x}{x} \; dx.$$

Now by the substitution

$$x = \frac{u^2-1}{2u} \quad \Longleftrightarrow \quad u = x + \sqrt{x^2+1},$$

and the easy equality $\sinh^{-1} y = \log \left( y + \sqrt{y^2+1} \right)$, we have

$$\int_{0}^{y} \frac{\sinh^{-1}x}{x} \; dx = -\int_{1}^{y + \sqrt{y^2+1}} \left( \frac{2u}{1-u^2} + \frac{1}{u} \right) \log u \; du = -F\bigg(y+\sqrt{y^2+1}\bigg),$$

where

$$F(s) := \int_{1}^{s} \left( \frac{2u}{1-u^2} + \frac{1}{u} \right) \log u \; du.$$

Now simple observation shows that for $s > 0$ we have

$$F\left( \frac{1}{s} \right) = -F(s).$$

Since $y + \sqrt{y^2+1} \gg 1$ whenever $y \gg 1$, in view of the identity above, we may calculate $F(s)$ for $s = \left( y + \sqrt{y^2+1} \right)^{-1} \in (0, 1)$ instead since

$$F\left(y+\sqrt{y^2+1}\right) = -F\left(\frac{1}{y+\sqrt{y^2+1}}\right) = -F(s) .$$

Now we introduce the dilogarithm function, defined by

$$\mathrm{Li}_{2} (x) = \sum_{n=1}^{\infty} \frac{x^n}{n^2} = -\int_{0}^{x} \frac{\log(1-t)}{t} \; dt.$$

Then

\begin{align*} F(s) &= \frac{1}{2} \int_{1}^{s} \frac{\log(u^2)}{1-u^2} \; (2udu) + \int_{1}^{s} \frac{\log u}{u} \; du \\ &= \frac{1}{2} \int_{1}^{s^2} \frac{\log v}{1-v} \; dv + \frac{1}{2} \log^2 s \qquad (v = u^2) \\ &= - \frac{1}{2} \int_{0}^{1-s^2} \frac{\log (1-w)}{w} \; dw + \frac{1}{2} \log^2 s \qquad (w = 1-v) \\ &= \frac{1}{2} \mathrm{Li}_{2}(1-s^2) + \frac{1}{2} \log^2 s. \qquad (w = 1-v) \end{align*}

Thus plugging back, we have

$$\int_{0}^{y} \frac{\sinh^{-1}x}{x} \; dx = \frac{1}{2} \left[ \mathrm{Li}_{2} \left( \frac{2y}{y+\sqrt{y^2+1}}\right) + \log^2 \left(y+\sqrt{y^2+1}\right) \right].$$

This shows that

\begin{align*} \log^2 y - 2\int_{0}^{y} \frac{\log x}{\sqrt{x^2+1}}\;dx &= \log^2 y - 2 \log y \, \log \left( y + \sqrt{y^2+1} \right) \\ & \qquad + \mathrm{Li}_{2} \left( \frac{2y}{y+\sqrt{y^2+1}}\right) + \log^2 \left(y+\sqrt{y^2+1}\right) \\ &= \mathrm{Li}_{2} \left( \frac{2y}{y+\sqrt{y^2+1}}\right) + \log^2 \left(1+\sqrt{1+y^{-2}}\right), \end{align*}

which clearly converges to

$$\mathrm{Li}_{2}(1) + \log^2 2 = \zeta(2) + \log^2 2 = \frac{\pi^2}{6} + \log^2 2.$$

Numerical experiment shows that this converges to its limit relatively fast.

In fact, reflection formula for dilagorithm gives the following estimate.

$$\log^2 y - 2\int_{0}^{y} \frac{\log x}{\sqrt{x^2+1}}\;dx = \frac{\pi^2}{6} + \log^2 2 - \frac{\log y}{2y^2} + O\left(\frac{1}{y^2}\right).$$

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Perfect answer. – nbubis Jun 29 '12 at 16:12
Thank you for your praise. – Sangchul Lee Jun 29 '12 at 16:19
Thank you for this work! – Martin Gales Jul 2 '12 at 4:55

An alternative approach: Write $\ln^2y$ as the integral of its derivative: $$\ln^2 y = 2\int_1^y \frac{\ln x}{x}dx,$$ and then combine with the integral that explicitly appears in $F(y)$, where $F$ is the function whose limit you want to calculate. Combining yields $$F(y) = 2\int_1^y\frac{\ln x}{x\sqrt{x^2+1}(x+\sqrt{x^2+1})}dx - 2\int_0^1\frac{\ln x}{\sqrt{x^2+1}}dx,$$ and so the limit is $$L = 2\int_1^\infty\frac{\ln x}{x\sqrt{x^2+1}(x+\sqrt{x^2+1})}dx - 2\int_0^1\frac{\ln x}{\sqrt{x^2+1}}dx,$$ where the improper infinite integral clearly converges due to $\ln x/x^3$ fall-off as $x\to\infty$. This at least shows that the limit exists. As for evaluating those integrals...Mathematica yields a bunch of log, polylog and arcsinh terms eventually yielding the same result that @sos440 found in a much nicer analytic way.

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Yes, this is a good point! Thank you. – Martin Gales Jul 2 '12 at 5:03

Take $y=e^{1/t}$,then this limit becomes $$\lim_{t\to0^+}\frac{1-2t^2\int_0^{e^{1/t}}\frac{\ln x}{\sqrt{x^2+1}}dx}{t^2}$$.Now you can solve it easily using L' Hopital's Rule.

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You already have the answer to this? – Martin Gales Jun 29 '12 at 15:24

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The Question: $\ds{\lim_{y \to \infty}\bracks{\ln^{2}\pars{y} - 2\int_{0}^{y}{\ln\pars{x}\over \root{x^{2} + 1}}\,\dd x} =\, ?}$

\begin{align} &\color{#f00}{\lim_{y \to \infty}\bracks{\ln^{2}\pars{y} - 2\int_{0}^{y}{\ln\pars{x}\over \root{x^{2} + 1}}\,\dd x}} = \lim_{y \to \infty}\bracks{\ln^{2}\pars{y} - \int_{x = 0}^{x = y} {x \over \root{x^{2} + 1}}\,\dd\bracks{\ln^{2}\pars{x}}} \\[3mm] = & \int_{0}^{\infty}{\ln^{2}\pars{x} \over \pars{x^{2} + 1}^{3/2}}\,\dd x = \lim_{\mu \to 0}\partiald[2]{}{\mu} \int_{0}^{\infty}{x^{\mu} \over \pars{x^{2} + 1}^{3/2}}\,\dd x\ \stackrel{x^{2}\ \to x}{=}\ \half\,\lim_{\mu \to 0}\partiald[2]{}{\mu} \int_{0}^{\infty}{x^{\mu/2 - 1/2} \over \pars{x + 1}^{3/2}}\,\dd x \end{align}

With the identity $\ds{\int_{0}^{\infty}{t^{a - 1} \over \pars{1 + t}^{a + b}}\,\dd t = {\Gamma\pars{a}\Gamma\pars{b} \over \Gamma\pars{a + b}}}$: \begin{align} &\color{#f00}{\lim_{y \to \infty}\bracks{\ln^{2}\pars{y} - 2\int_{0}^{y}{\ln\pars{x}\over \root{x^{2} + 1}}\,\dd x}} = \half\,\lim_{\mu \to 0}\partiald[2]{}{\mu}\bracks{% {\Gamma\pars{\mu/2 + 1/2}\Gamma\pars{1 - \mu/2} \over \Gamma\pars{3/2}}} \\[3mm] = &\ \color{#f00}{{\pi^{2} \over 6} + \ln^{2}\pars{2}} \approx 2.1254 \end{align}

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