# Prove that: $\int_{0}^{1} \frac{x^{4}\log x}{x^2-1}\le \frac{1}{8}$

Here is another interesting integral inequality :

$$\int_{0}^{1} \frac{x^{4}\log x}{x^2-1}\le \frac{1}{8}$$

According to W|A the difference between RS and LS is extremely small, namely 0.00241056. I don't know what would work here since the difference is so small.

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If there will be badges for those who asks interesting question, surely you will get it :) – Norbert Jun 22 '12 at 18:38
Thank you all for the solutions you gave! (+1) for all – user 1618033 Jun 22 '12 at 19:59
@Norbert: well, it's enough for me if people here enjoy my questions. :-) – user 1618033 Jun 22 '12 at 20:02

You can actually just evaluate the integral explicitly. You can divide $x^2 -1$ into $x^4$ and get $$\frac{x^4}{x^2 - 1} = x^2 + 1 + \frac {1}{x^2 - 1}$$ So the integral is the same as $$\int_0^1 (x^2 + 1)\log(x)\,dx + \int_0^1 \frac{\log(x)}{x^2 - 1}\,dx$$ The second integral is related to the famous dilogarithm integral, and as explained in Peter Tamaroff's answer can be evaluated to $\frac{\pi^2}{8}$. For the first term, just integrate by parts; you get $$({x^3 \over 3} + x)\log(x)\big|_{x = 0}^{x =1} - \int_0^1 ({x^2 \over 3} + 1)\,dx$$ The first term vanishes, while the second term is $-{10 \over 9}$. So the answer is just ${\pi^2 \over 8} - {10 \over 9}$ which is less than ${1 \over 8}$.

A way of doing the whole integral in one fell swoop occurs to me. Note that ${\displaystyle {1 \over 1 - x^2} = \sum_{n=0}^{\infty} x^{2n}}$. So the integral is $$-\sum_{n = 0}^{\infty} \int_0^1 x^{2n + 4}\log(x)\,dx$$ $$= -\sum_{m = 2}^{\infty} \int_0^1 x^{2m}\log(x)\,dx$$ Integrating this by parts this becomes $$\sum_{m = 2}^{\infty} \int_0^1 {x^{2m} \over 2m + 1}$$ $$= \sum_{m = 2}^{\infty} {1 \over (2m + 1)^2}$$ This is the sum of the reciprocals of the odd squares starting with $5$. The sum of the reciprocals of all odd squares is ${\pi^2 \over 8}$, so one subtracts off $1 + {1 \over 9} = {10 \over 9}$. Hence the result is ${\pi^2 \over 8} - {10 \over 9}$.

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Haha, I just went straight to an inequality! Silly me! – Pedro Tamaroff Jun 22 '12 at 19:52
@Zarrax: that's a really interesting approach! – user 1618033 Jun 22 '12 at 20:01
Practically this solution uses @PeterTamaroff work, without it you could as well write right away $\frac{\pi^2}{8} - \frac{10}{9} \le \frac{1}{8}$ :/ – qoqosz Jun 22 '12 at 20:07
@qoqosz Of course what he did there was the bulk of the effort... the point is, once you have that you can just do the original integral. – Zarrax Jun 22 '12 at 20:12
$+\left\lceil \frac 12 \right\rceil$ for the second part. – Sam Jun 28 '12 at 21:34

Representing natural logarithm as an integral and changing the order of integration we obtain: $$\ldots = \int_0^1 \frac{x^4}{x^2 - 1} \, dx \int_1^x \frac{dt}{t} = \int_0^1 \frac{dt}{t} \int_0^t \frac{x^4}{x^2 - 1} \, dx \\= \int_0^1 \frac{t + \frac{1}{3} t^3 - \tanh^{-1} t}{t}\, dt = \frac{10}{9} - \int_0^1 \frac{\tanh^{-1} t}{t} \, dt$$ So we now want to show that: $$\frac{71}{72} = \frac{10}{9} - \frac{1}{8} \le \int_0^1 \frac{\tanh^{-1} t}{t} \, dt$$ Using Maclauirn series we have: $$\frac{\tanh^{-1}}{t} = 1 + \frac{t^2}{3} + \frac{t^4}{5} + \ldots$$ Integrating first three terms yields: $\frac{259}{225} > \frac{71}{72}$.

Actually it turns out that $\frac{\tanh^{-1} t}{t} \approx 1$ is enough.

Alternatively we can write: $$\int_0^1 \frac{\tanh^{-1} t}{t} dt = \int_0^1 \frac{dt}{t} \int_0^t \frac{dx}{1-x^2} \ge \int_0^1 \frac{dt}{t} \int_0^t dx = \int_0^1 dt = 1 \ge \frac{71}{72}$$

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Maybe this is easier:

\eqalign{ & \int\limits_0^1 {\frac{{{x^4}\log x}}{{{x^2} - 1}}dx} = \int\limits_0^1 {\frac{{\left( {{x^4} - 1} \right)\log x}}{{{x^2} - 1}}dx} + \int\limits_0^1 {\frac{{\log x}}{{{x^2} - 1}}dx} \cr & = \int\limits_0^1 {\left( {{x^2} + 1} \right)\log xdx} + \int\limits_0^1 {\frac{{\log x}}{{{x^2} - 1}}dx} \cr}

$$= \int\limits_0^1 {\left( {{x^2} + 1} \right)\log xdx} = \left[ {\left( {\frac{{{x^3}}}{3} + x} \right)\log x} \right]_0^1 - \int\limits_0^1 {\left( {\frac{{{x^2}}}{3} + 1} \right)dx} = - \frac{{10}}{9}$$

$$\int_0^1 \frac{\log x}{x^2-1}dx=\frac 1 2 \int_0^1 \frac{\log x}{x-1}dx-\frac 1 2 \int_0^1 \frac{\log x}{x+1}dx$$

Now those two evaluate in terms of $\pi$, since we get

$$-\int_0^1 \frac{\log x}{1-x}dx=-\int_0^1 \log x \sum_{k=0}^\infty x^kdx=$$

$$=-\sum_{k=0}^\infty\int_0^1 x^k\log x dx=\sum_{k=0}\frac{1}{(k+1)^2}=\frac{\pi^2}{6}$$

Since

$$\int_0^1 x^k\log x dx=-\frac{1}{(k+1)^2}$$

And for the other, we get the similar:

$$\int_0^1 \frac{\log x}{1+x}dx=\int_0^1 \log x \sum_{k=0}^\infty (-1)^kx^kdx=$$

$$=\sum_{k=0}^\infty\int_0^1 \log x (-1)^kx^kdx=\sum_{k=0}^\infty(-1)^k\int_0^1 x^k\log x dx=\sum_{k=0}\frac{(-1)^{k+1}}{(k+1)^2}=-\frac{\pi^2}{12}$$

So we have that

$$\int_0^1 \frac{\log x}{x^2-1}dx=\frac 1 2 \left( \frac{\pi^2}{6}+\frac{\pi^2}{12} \right)$$

$$\int_0^1 \frac{\log x}{x^2-1}dx=\frac{\pi^2}{8}$$

and finally

$$I=\frac{\pi^2}{8}-\frac{10}{9}\approx 0.1225 < 0.125$$

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it's a bit longer proof, but it's a detailed answer. That's a good point! Thanks. – user 1618033 Jun 22 '12 at 20:20
You can do the $\int_0^1 {\log(x) \over x^2 - 1}$ integral the same way you did the two integrals there, expand ${1 \over 1 - x^2} = \sum_n x^{2n}$. You end out with the sum of the reciprocals of the odd squares, which gives $\frac {\pi^2}{8}$. – Zarrax Jun 22 '12 at 20:51
@Zarrax Hmhm. True. – Pedro Tamaroff Jun 22 '12 at 21:07
@Peter Tamaroff: i will delete the above posts. ;) – user 1618033 Jun 28 '12 at 21:19

According to Maple, your integral is $\dfrac{\pi^2}{8} - \dfrac{10}{9}$, so your inequality becomes $\pi < \sqrt{89}/{3}$. In fact, an antiderivative is $$F(x) = \dfrac{x^3 \ln(x)}{3} - \dfrac{x^3}{9} + x \ln(x) - x - \dfrac{\ln(x) \ln(x+1)}{2} - \dfrac{\text{dilog}(x)+\text{dilog}(x+1)}{2}$$

More generally, for $p > -1$ $$\int_0^1 \dfrac{x^p \ln(x)}{x^2-1}\ dx = \dfrac{\Psi(1,(p+1)/2)}{4}$$ where for even integers $p=2n$, $$\Psi(1,n+1/2) = \sum_{k=n+1}^\infty \dfrac{4}{(2k-1)^2}$$ while for odd integers $p=2n-1$, $$\Psi(1,n) = \sum_{k=n}^{\infty} \dfrac{1}{k^2}$$

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hmmm. That's interesting. – user 1618033 Jun 22 '12 at 18:02