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

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.

Update: The approach used by achille hui in a series of answers (see 1, 2) maybe can be helpful, even if I'm not sure about it.

• It seems that $$\int_{0}^{1}\frac{\log\left(x\right)}{\sqrt{1+x^{4}}}dx=\frac{1}{e^{-i \pi/4}}\int_0^1 \frac{F\left(\left.i \sinh ^{-1}\left(y\, e^{i \pi /4}\right)\right|-1\right)}{y} \, dy$$ where $F(\phi |m)$ is elliptic integral of the first kind. I don't understand (using a substitution for example) how this integral form can be transformed into an integral of the form you desire. May 13, 2021 at 23:12
• $$\int_0^1 \frac{\log (x)}{\sqrt{x^4+1}} \, dx=-\int_0^1 \int_0^1 \frac{1}{\sqrt{1+x^4 y^4}} \, dx\, dy$$ May 13, 2021 at 23:20
• From Mathematica (may simplify to something else) $$\int_0^1 \frac{\log (x)}{\sqrt{x^4+1}} \, dx=\frac{1}{3 \sqrt{2}}+\frac{2 \Gamma \left(\frac{5}{4}\right)^2}{\sqrt{\pi }}+\frac{4}{3} e^{i \pi/ 4}\, F\left(\left.i \sinh ^{-1}\left(e^{i \pi/ 4}\right)\right|-1\right)-\frac{1}{6} \, _3F_2\left(-\frac{3}{4},-\frac{3}{4},-\frac{1}{2};\frac{1}{4},\frac{1}{4};-1\right)$$ May 13, 2021 at 23:30
• Let $f$ be a certain Eisenstein form of principal congruence subgroup $\Gamma(4)$, then the integral is equivalent to the value at $i$ of a certain 3-rd primitive $g$ of $f$ (i.e. $g''' = f$). We know next to nothing regarding special values of these functions, except some miraculous evaluations arising directly from functional equation. Such miracle, does not happen for our $g$. So I am pessimistic on this approach. May 15, 2021 at 19:23
• Another perspective: the integral is equivalent to $$\int_0^{\omega/4} \log \wp(x) dx \qquad g_2 = -1, g_3 = 0, \omega\text{ the real period of }\wp$$ this in term boils down to computing some values of $F(z) = \int_0^z \log \sigma(z) dz$ for the Weierstrass sigma. Note that $F'''(z) = -\wp(z)$, one can calculate 1st or 2nd integral of $\wp(z)$ at torsion point, but for 3rd integral $F$, I am out of idea. $F(z)$ is intractable except at lattice or $2$-torsion points (corresponds to easy integrals like $\int_0^\infty \log x /\sqrt{1+x^4} dx$), but the problem is about $4$-torsion. May 15, 2021 at 19:39

Long Comment:

I found that you can approximate the original integral by the following process of repeated integration

$$\int_0^1 \frac{\log (x)}{\sqrt{x^4+1}} \, dx=-\frac{2 \Gamma \left(\frac{5}{4}\right)^2}{\sqrt{\pi }}+2 \int_0^1 \frac{x^4 \log (x)}{\left(x^4+1\right)^{3/2}} \, dx$$

$$2 \int_0^1 \frac{x^4 \log (x)}{\left(x^4+1\right)^{3/2}} \, dx=\frac{1}{5 \sqrt{2}}-\frac{2 \Gamma \left(\frac{5}{4}\right)^2}{5 \sqrt{\pi }}+\frac{12}{5}\int _0^1\frac{ x^8 \log (x)}{\left(x^4+1\right)^{5/2}}$$

$$\frac{12}{5} \int_0^1 \frac{x^8 \log (x)}{\left(x^4+1\right)^{5/2}} \, dx=\frac{2}{15\sqrt{2}}-\frac{2 \Gamma \left(\frac{5}{4}\right)^2}{9 \sqrt{\pi }}+\frac{24}{9} \int_0^1 \frac{x^{12} \log (x)}{\left(x^4+1\right)^{7/2}} \, dx$$

$$\frac{24}{9} \int_0^1 \frac{x^{12} \log (x)}{\left(x^4+1\right)^{7/2}} \, dx=\frac{19}{195 \sqrt{2}}-\frac{2 \Gamma \left(\frac{5}{4}\right)^2}{13 \sqrt{\pi }}+\frac{112}{39} \int_0^1 \frac{x^{16} \log (x)}{\left(x^4+1\right)^{9/2}} \, dx$$

After four iterations above, it is found that

$$\int_0^1 \frac{\log (x)}{\sqrt{x^4+1}} \, dx\approx \frac{14 \sqrt{2}}{65}-\frac{1624 \Gamma \left(\frac{5}{4}\right)^2}{585 \sqrt{\pi }}$$

with an error of less than $$0.13$$ %

Update 05/01/2022 - Playing around tonight I've found further conjectural simplifications to update comments above, but no closed form in terms of special functions:

$$\int_0^1 \frac{\log (x)}{\sqrt{x^4+1}} \, dx=\frac{1}{6} \left(-\, _3F_2\left(-\frac{3}{4},-\frac{3}{4},-\frac{1}{2};\frac{1}{4},\frac{1}{4};-1\right)+\sqrt{2}-\frac{4 \Gamma \left(\frac{5}{4}\right)^2}{\sqrt{\pi }}\right)\tag{1}$$

$$\int_0^1 \frac{\log (x)}{\sqrt{x^4+1}} \, dx=\frac{1}{3} \left(\, _3F_2\left(-\frac{3}{4},-\frac{3}{4},\frac{1}{2};\frac{1}{4},\frac{1}{4};-1\right)-\sqrt{2}+\frac{4 \Gamma \left(\frac{5}{4}\right)^2}{\sqrt{\pi }}\right)\tag{2}$$

$$\frac{4}{3} e^{i \pi/ 4}\, F\left(\left.i \sinh ^{-1}\left(e^{i \pi/ 4}\right)\right|-1\right)=-\frac{8 \Gamma \left(\frac{5}{4}\right)^2}{3 \sqrt{\pi }}$$

where $$F(\phi |m)$$ is elliptic integral of the first kind.

Assuming $$(1)$$ and $$(2)$$ are correct then subtracting one from the other gives: $$2 \, _3F_2\left(-\frac{3}{4},-\frac{3}{4},\frac{1}{2};\frac{1}{4},\frac{1}{4};-1\right)+\, _3F_2\left(-\frac{3}{4},-\frac{3}{4},-\frac{1}{2};\frac{1}{4},\frac{1}{4};-1\right)=3 \sqrt{2}-\frac{12 \Gamma \left(\frac{5}{4}\right)^2}{\sqrt{\pi }}$$

Update 17/01/2022 - Another Mathematica derived series approximation that is more convenient to use in a CAS, yet related to my original observation is

$$\int_0^1 \frac{\log (x)}{\sqrt{x^4+1}} \, dx=-\underset{m\to \infty }{\text{lim}}\int_0^1 \left(\sum _{n=0}^m \frac{(2 n)!\, x^{4 n}}{n! \, \left(x^4+1\right)^{\frac{1}{2} (2 n+1)} \left(\prod _{k=0}^n (4 k+1)\right)}\right) \, dx$$

and $$\sum _{n=0}^\infty \frac{(2 n)!\, x^{4 n}}{n! \, \left(x^4+1\right)^{\frac{1}{2} (2 n+1)} \left(\prod _{k=0}^n (4 k+1)\right)} \,=\frac{\, _2F_1\left(\frac{1}{2},1;\frac{5}{4};\frac{x^4}{x^4+1}\right)}{\sqrt{x^4+1}}$$

• “05/01/2022”? That’s in the future. Lol Jan 9, 2022 at 0:03
• @RadialArmSaw en.wikipedia.org/wiki/Calendar_date Jan 12, 2022 at 9:33

$$I=-\int\limits_0^\infty \dfrac{x e^{-x}}{\sqrt{1+e^{-4x}}}\,\text dx =-\int\limits_0^\infty \dfrac{x}{\sqrt{e^{2x}+e^{-2x}}}\,\text dx =-\dfrac1{\sqrt2}\int\limits_0^\infty \dfrac{x}{\sqrt{2\cosh^2x-1}}\,\text dx,$$ $$I=-\dfrac1{2}\int\limits_0^\infty \dfrac{x}{\cosh x\sqrt{1-\dfrac1{2\cosh^2x}}}\,\text dx =\sum\limits_{k=0}^\infty \dfrac1{(-2)^{k+1}}\dbinom{-^1\!/_4}kI_k,\tag1$$ where $$I_k=\int\limits_0^\infty \dfrac{x}{\cosh^{2k+1}x}\,\text dx.\tag2$$ Then $$\begin{cases} I_0=2C\approx1.831931184,\quad I_1=C-\dfrac12\approx0.415965594,\\[5pt] I_2=\dfrac{3}{4}C-\dfrac{11}{24}\approx 0.228640862,\quad I_3=\dfrac58C-\dfrac{299}{720} \approx 0.157200719,\\[6pt] I_4=\dfrac{35}{64}C-\dfrac{15371}{40320} \approx 0.119693486,\quad I_5=\dfrac{63}{128}C+\dfrac{142819}{403200} \approx 0.096613026\dots, \tag3 \end{cases}$$ where $$\;C\;$$ is the Catalan constant.

At the same time, $$I_{k+1}=I_{k}-\int\limits_0^\infty \dfrac{x \tanh^2 x\,\text dx}{\cosh^{2k+1}x} =I_k+\int\limits_0^\infty \dfrac{x\sinh x}{\cosh^{2k+1}x}\,\text d\,\dfrac1{\cosh x}$$ $$=I_k-\int\limits_0^\infty \dfrac{-(2k+1)\sinh^2 x+\cosh^2 x}{\cosh^{2k+3}x}\,x\,\text dx-\dfrac1{2k+1}\,\dfrac1{\cosh^{2k+1}x}\bigg|_0^\infty$$ $$=I_k-(2k+1)I_{k+1}+2kI_k+\dfrac1{2k+1},$$ $$I_{k+1}=\dfrac{2k+1}{2k+2} I_k-\dfrac{1}{(2k+1)(2k+2)}.\tag4$$

Obtained series can converge faster than the closed form from OP.

Seems the given formula (1) is oversimplified for a mathematical understanding.

It holds by simply making use of the argument simplification rule for the Hypergeometric Function $$_{2}F_{1}$$:

$$(1-z)^{a+b-c}_{2}F_{1}(a,b,c,,z)=_{2}F_{1}(-a+c-b+c,c,z)$$

with

$$z=-x^4$$.

Making then the integration:

$$\int \frac{log(x)}{\sqrt{1+x^4}}=x$$ $$_{2}F_{1}(\frac{1}{4},\frac{1}{2};\frac{5}{4};-x^4) log(x) - x$$ $$_{3}F_{2}(\frac{1}{4},\frac{1}{4},\frac{1}{2};\frac{5}{4},\frac{5}{4};-x^4) + constant$$

$$_{3}F_{2}$$ is the generalized hypergeometric function.

This goes over to the given result is the appropriate borders of integrations are entered.

This can be confirmed by modern CAS or formula collection for the two types of hypergeometric functions. For example, the integral definitions can be found on HypergeometricPFQ.

More special for the question is this definition: Hypergeometric3F2. Useful is just the integral representations in the section for both hypergeometric functions.

For the numerical value and some impressive representations look at

HypergeometricPFQ[{0.25,0.25,0.5},{1.25,1.25},-1]

The numerical value is $$-0.983384$$.

So proving the formula (1) is using integral identities plainly and integration method for $$log$$ production integrations. Then this is only using formulas for coefficients for both hypergeometric functions. But calculating the value is much more difficult and a bit too far beyond a simple answer. This symbolical representation is higher complex functions representation theory and methodology:

.

Have a look at the formula collection mentioned above.