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.

Solution 1:

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} $.

Solution 2:

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.

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