# About the integral $\int_{0}^{1}\frac{\log(x)}{\sqrt{1+x^{4}}}dx$ and elliptic functions

NOTE: I post this question on math.stackexchange but nobody answered, so I try here.

For a work we need to evaluate the following integral $$\int_{0}^{1}\frac{\log\left(x\right)}{\sqrt{1+x^{4}}}dx=\,-_{3}F_{2}\left(\frac{1}{4},\frac{1}{4},\frac{1}{2};\frac{5}{4},\frac{5}{4};-1\right).\tag{1}$$ Classical approaches seem to lead nowhere, but it is possible to translating the problem into the language of elliptic functions. Let $$\text{sn}(u,k)$$ be the Jacobi elliptic sine. We can prove that the evaluation of $$(1)$$ boils down to the evaluation of $$\int_{0}^{T/4}\log\left(-e^{-\pi i/4}\text{sn}\left(e^{3\pi i/4}z,-1\right)\right)dz$$ where $$T=2K(1/2)$$ and $$K\equiv K(k)$$ is the complete elliptic integral of the first kind with $$k$$ the elliptic modulus. I am not an expert in elliptic functions so I have difficulty to understand if this integral can be evaluated or not. However, I found this formula $$\log\left(\text{sn}\left(u,k\right)\right)=\log\left(\frac{2K}{\pi}\right)+\log\left(\sin\left(\frac{\pi u}{2K}\right)\right)-4\sum_{n\geq1}\frac{1}{n}\frac{q^{n}}{1+q^{n}}\sin^{2}\left(\frac{n\pi u}{2K}\right)$$ with $$q\equiv e^{-\pi\frac{K}{K^{\prime}}}=e^{\pi i\tau}$$ and $$\left|\text{Im}\left(\frac{\pi u}{2K}\right)\right|<\frac{\pi}{2}\text{Im}\left(\tau\right)$$. So, assuming that we can exchange the integral with the series, which I'm not sure about, the problem boils down to studying the following Lambert series $$\sum_{n\geq1}\frac{1}{n^{2}}\frac{q^{n}}{1+q^{n}}\sin\left(\frac{\pi nT}{4K}\right).\tag{2}$$ I have seen that similar series have been studied but this particular one has not (as far as I know). Clearly, there are a lot of heuristic passages and so I may have written nonsense.

Questions:

$$1)$$ Is it possible to find a closed form (in terms of special functions) of $$(1)$$?

$$2)$$ Assuming that the “elliptic approach” is correct, is there a closed form of $$(2)$$, maybe in terms of elliptic functions?

Thank you

• I think 2 days is not long enough to conclude no answers will be received. May 12 at 15:21
• Equivalently, it's $$\frac{\sqrt{2}}8 \int_{-\infty}^0 \frac{t{\rm d}t}{\sqrt{\cosh(t)}}.$$ May 12 at 20:51
• It probably doesn't help, but one can rewrite your integral in a more conceptual form. Namely, working on the elliptic curve $E:y^2=x^4+1$, the function $\log(x)$ is the indefinite integral of the differential form $\frac{dx}{x}$. Then your integral should be the iterated integral $\int_\gamma \frac{dx}{y} \frac{dx}{x}$ with $\gamma : (0,1) \to (1,\sqrt{2})$. There is a whole theory of iterated integrals, for the projective line you get polylogarithms. So here you have some kind of elliptic polylog, although I don't know if the differential forms above are natural enough. May 14 at 15:15
• You could try Levin, Elliptic polylogarithms. General Theory and Applications, hse.ru/data/2020/03/17/1525050020/Summary.pdf and Brown-Levin, Multiple Elliptic Polylogarithms, arxiv.org/abs/1110.6917 May 14 at 15:18
• math.stackexchange.com/questions/570997/… Jul 19 at 13:36

If you only care just about that one definite integral, I don't see why you would need some "closed form" expression for it. After all, the integral itself can be efficiently evaluated to arbitrary precision. For example, here are the first 200 or so digits:

-0.98338406775370959402527563848963167639494869596755320855722733745669900479910600590043880520941926876314752117646474005285439505408063547659244909334265635752359407651607735392461654885991979078790988964771001196...

I evaluated it (from the form given by Max Alekseyev) in SAGE, which uses the Arb C library for efficient high precision integration of analytic functions.

• I think it's hard to explain in a comment, but essentially this integral is part of a larger structure, so to speak, and I need to understand if it can be written in terms of some special function. However, thank you for your answer. May 13 at 7:36

$$\int_{0}^{1}\frac{\ln x}{\sqrt{1+x^{4}}}\,dx=-\sum _{n=0}^{\infty } \binom{2 n}{n} \frac{(-1)^n}{(4 n+1)^2 4^n}$$
the $$(-1)^n$$ spoils things, without that factor the right-hand-side would evaluate to $$\pi ^{3/2} \Gamma (5/4)/\,\Gamma (-1/4)$$.
• Indeed that $(-1)^{n}$ creates a lot of problems and I don't know how to "get around" it. May 12 at 15:42
• Alternatively, the $(-1)^n$ doesn't spoil things at all, it makes it so much better for convergence: 6 terms only for less than 1% error (rate of convergence of alternating series..). Does the OP need a more precise evaluation? May 12 at 20:11