Find the integral $\int_0^1 \frac{x^2 - 1}{ \ln(x)} dx$

Question

I tried substituting $\ln (x)$ as $t$, but it led to no standard integral for me to move further.

Substituting $\ln (x)$ as $-t$ gives $$ \int_0^\infty\frac{e^{-t}-e^{-3t}}{t}\,\mathrm{d}t $$

@robjohn because it's not an elementary integral. It's a special integral and I'm pretty sure they didn't teach me that for high school — Harsha G., Sep 21 '17 at 14:41
The Frullani integral only uses that $\int_0^\infty f(x),\mathrm{d}x=\lim\limits_{\epsilon\to0^+}\int_\epsilon^\infty f(x),\mathrm{d}x$ and $e^x$ is continuous at $x=0$. I will add an answer that expands on that. — robjohn, Sep 21 '17 at 15:01

Robert Z · Answer 1 · 2017-09-21T14:17:19.963

6

In general, for $a>-1$, by letting $t=-\ln(x)$, we obtain $$\int_0^1 \frac{x^a-1}{\ln x} dx=\int_0^{+\infty} \frac{e^{-t}-e^{-(a+1)t}}{t} dt=\ln(a+1)$$ where in the last step we used a known result about Frullani's integral with $f(t)=e^{-t}$.

edited Sep 21 '17 at 14:17

answered Sep 13 '17 at 06:12

Robert Z

145,942

How do you know that the integral converges? Or can we always use the formula when it gives a real number? Also, can we generalize to more $a$ than $a>0$? $\ln(a+1)$ is still a real number if $a>-1$. – user263326 Sep 21 '17 at 13:29
It seems that according to this paper given in another comment, $a>-1$ may also be possible (see Example 3.1.). – user263326 Sep 21 '17 at 13:33
@user263326 Yes, you are right, it converges for $a>-1$. In fact, near $0^+$ the integrand is $\sim 1/(x^{-a}\ln(x)$ whose integral is convergent when $-a<1$. – Robert Z Sep 21 '17 at 14:16

score 5 · Answer 2 · answered Sep 19 '17 at 18:28

$\newcommand{\bbx}[1]{\,\bbox[15px,border:1px groove navy]{\displaystyle{#1}}\,} \newcommand{\braces}[1]{\left\lbrace\,{#1}\,\right\rbrace} \newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack} \newcommand{\dd}{\mathrm{d}} \newcommand{\ds}[1]{\displaystyle{#1}} \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,} \newcommand{\ic}{\mathrm{i}} \newcommand{\mc}[1]{\mathcal{#1}} \newcommand{\mrm}[1]{\mathrm{#1}} \newcommand{\pars}[1]{\left(\,{#1}\,\right)} \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,} \newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}} \newcommand{\verts}[1]{\left\vert\,{#1}\,\right\vert}$ \begin{align} \int_{0}^{1}{x^{2} - 1 \over \ln\pars{x}}\,\dd x & = \int_{0}^{1}\pars{x + 1}\ \overbrace{{x - 1 \over \ln\pars{x}}}^{\ds{= \int_{0}^{1}x^{t}\,\dd t}}\ \,\dd x = \int_{0}^{1}\int_{0}^{1}\pars{x^{t + 1} + x^{t}}\,\dd x\,\dd t \\[5mm] & = \int_{0}^{1}\pars{{1 \over t + 2} + {1 \over t + 1}}\,\dd t = \bracks{\ln\pars{3} - \ln\pars{2}} + \ln\pars{2} \\[5mm] & = \bbx{\ln\pars{3}} \approx 1.099 \end{align}

(+1) Clever, but does require noticing the fact about $\frac{x-1}{\log(x)}$ — robjohn, Sep 21 '17 at 14:33
@robjohn Integrals $\int_{0}^{1}x^{t}t^{n},\mathrm{d}t$ and $\int_{0}^{\infty}x^{t}t^{n},\mathrm{d}t$ yield interesting identities for suitable values of $n$ and $x$. Thanks. — Felix Marin, Sep 21 '17 at 17:18

score 4 · Accepted Answer · answered Sep 13 '17 at 05:55

4

Hint: for $a>0$, write $$f(a) = \int_0^1 \frac{x^a-1}{\ln x} dx$$ Then we can differentiate with respect to $a$ to give $f'(a) = 1/(a+1)$. Also note that $$\lim_{a\to 0^+} f(a) = 0$$ This can help you to find the integration constant.

answered Sep 13 '17 at 05:55

pisco

18,983

Hmm I have never seen this kind of method before. What is this called? We introduced a new variable.. when can I use this? – Harsha G. Sep 13 '17 at 06:02
It is called "differentiation under the integral", there are some examples in wikipedia. – pisco Sep 13 '17 at 06:09
2

@Harsha G. Take a look at this article Example 3.1 will be your integral. – Sep 13 '17 at 06:13
read http://math.stackexchange.com/questions/1505423/evaluation-of-int-01-fracx2015-1-ln-xdx and http://math.stackexchange.com/questions/2017018/how-to-integrate-int-01-fracx7-1-logxdx and https://math.stackexchange.com/questions/2239530/using-feymanns-trick-in-advanced-integration – FDP Sep 13 '17 at 10:01

robjohn · Answer 4 · 2017-09-21T15:21:50.160

Expanding on Robert Z's answer, $$ \begin{align} \int_0^1\frac{x^2-1}{\log(x)}\,\mathrm{d}x &=-\int_0^\infty\frac{e^{-2x}-1}{x}\,e^{-x}\,\mathrm{d}x\\ &=\int_0^\infty\frac{1-e^{-2x}}{x}\,e^{-x}\,\mathrm{d}x\\ &=\int_0^\infty\frac{e^{-x}-e^{-3x}}{x}\,\mathrm{d}x\\ &=\lim_{\epsilon\to0^+}\int_\epsilon^\infty\frac{e^{-x}-e^{-3x}}{x}\,\mathrm{d}x\\ &=\lim_{\epsilon\to0^+}\left(\int_\epsilon^\infty\frac{e^{-x}}{x}\,\mathrm{d}x-\int_\epsilon^\infty\frac{e^{-3x}}{x}\,\mathrm{d}x\right)\\ &=\lim_{\epsilon\to0^+}\left(\int_\epsilon^\infty\frac{e^{-x}}{x}\,\mathrm{d}x-\int_{3\epsilon}^\infty\frac{e^{-x}}{x}\,\mathrm{d}x\right)\\ &=\lim_{\epsilon\to0^+}\int_\epsilon^{3\epsilon}\frac{e^{-x}}{x}\,\mathrm{d}x\\ &=\lim_{\epsilon\to0^+}\int_1^3\frac{e^{-\epsilon x}}{x}\,\mathrm{d}x\\ &=\lim_{\epsilon\to0^+}\int_1^3\left(\frac1x+\frac{o(\epsilon)}x\right)\,\mathrm{d}x\\ &=\lim_{\epsilon\to0^+}\left(\int_1^3\frac1x\,\mathrm{d}x+o(\epsilon)\right)\\ &=\int_1^3\frac1x\,\mathrm{d}x\\[9pt] &=\log(3) \end{align} $$

score 2 · Answer 5 · answered Sep 13 '17 at 06:11

The antiderivative is not elementary: $F(x) = \text{Ei}(1, -\ln(x)) - \text{Ei}(1,-3 \ln(x))$, where $$\text{Ei}(1,z) = \int_1^\infty e^{-tz} t^{-1}\; dt $$

As $x \to 0+$, $-\ln(x) \to +\infty$ and both $\text{Ei}(1, -\ln(x))$ and $\text{Ei}(1,-3 \ln(x))$ go to $0$. As $x \to 1-$, both $\text{Ei}(1,-\ln(x))$ and $\text{Ei}(1,-3 \ln(x)) \to \infty$, but $$\text{Ei}(1, 1/s) \sim \ln(s) -\gamma + O(1/s)$$ resulting in $\text{Ei}(1,-\ln(x)) - \text{Ei}(1,-3 \ln(x)) = \ln(3) + O(\ln(x))$. Thus the answer is $\ln(3)$.

Find the integral $\int_0^1 \frac{x^2 - 1}{ \ln(x)} dx$

5 Answers5