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I've been playing around with some integrals, and I started looking at Apery's constant.

There are some integral representation I've found online, such as:

$$\zeta(3)=\frac{16}{3}\int_{0}^{1}\frac{x\log^2(x)}{1+x^2}dx$$

and:

$$\zeta(3)=\frac{32}{7}\int_{0}^{1}\frac{x\log^2(x)}{1-x^4}dx$$

and these two monsters found on Wikipedia:

$$\zeta(3)=\frac{8\pi^2}{7}\int_{1}^{\infty}\frac{x(x^4-4x^2+1)\log\log x}{(1+x^2)^4}dx$$

$$\zeta(3)=\pi\int_{0}^{\infty}\frac{\cos(2\arctan x)}{(1+x^2)(\cosh \frac{\pi}{2}x)²}dx$$

Are there any other unique integrals that have this property? I would love to find other non-trivial ways of defining $\zeta(3)$.

Thanks in advance for any responses.

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3 Answers 3

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I do not know the reason of interest but here are few more: $$\begin{align} \zeta(3) &= \frac{1}{7}\int_0^\pi x(\pi-x)\csc(x) \,dx \tag{1}\\ &= \frac{4}{7}\int_{0}^{\pi/2} (4x-\pi)\ln(\sin x) \, dx \tag 1 \\ &=-\frac{1}{2}\int_0^1\int_0^1\frac{\ln(xy)}{1-xy}\,dx\,dy, \tag 2\end{align}$$

References: 1, 2.

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It should be noted that a "unique" integral is not an easy thing to define. This is not necessarily an answer but a clarification on the identities listed.

Let us start with the identity: $$ I(s) = \int_0^1 \frac{\ln^{s-1}(x)}{x} f(x) \: dx = (-1)^{s-1} \Gamma(s) \sum_{n=1}^{\infty}\frac{g(n)}{n^s} $$ where $f(x)$ is simply a power series with coefficients $g(k)$ $$ f(x) = \sum_{k=1}^{\infty} g(k) \: x^k $$

It can then be seen that your first "unique" integral has $f(x) = \frac{x^2}{1+x^2}$ in the above identity and the second integral has $f(x) = \frac{x^2}{1-x^4}$. These yield the sums $\sum _{n=1}^{\infty } \frac{(-1)^{n+1}}{(2 n)^3}$ and $\sum _{n=1}^{\infty } \frac{1}{(4 n - 2)^3}$, respectively.

The third integral can be derived from the derivative of the above identity: $$ \frac{d}{ds}I(s) = \int_0^1 \frac{\ln^{s-1}(x)}{x} \ln(\ln(x))f(x) \: dx = I(s)\cdot[\psi^{(0)}(s)+\pi i]+(-1)^{s} \Gamma(s) \sum_{n=1}^{\infty}\ln(n)\frac{g(n)}{n^s} $$ and noting that $$ \int_{1}^{\infty}\frac{x(x^4-4x^2+1)\ln (\ln(x))}{(1+x^2)^4}dx = \int_{0}^{1}\frac{x(x^4-4x^2+1)\ln(\ln(x))}{(1+x^2)^4}dx $$

The third integral from VIVID can also be derived this way as well. It is up to you if you wish to consider all of these integrals as "unique".

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Here is clean one $$\int_0^1 \frac{\ln^2(1-x^2)}xdx=\zeta(3)$$

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