Consider the terrible-horrible-not-good-very-bad integral $$I=\int_1^{\infty}\sqrt{\frac{\log x}{x^4+1}}dx$$ Where of course $\log x$ denotes the natural logarithm.


My friend asked me to evaluate $$\lim_{x\to1}\sqrt{\frac{\log x}{x^4+1}}$$ So I graphed it, which made me wonder if there was a closed-ish form for $$\int_1^{\infty}\sqrt{\frac{\log x}{x^4+1}}dx$$

I don't know where to even begin, because I can't think of any series that would give the integral. I'm sure the integrand doesn't have any elementary antiderivative, and I have no idea what an appropriate substitution for Feynman integration would be. I thought that it might be beneficial to try simplify it with a substitution of $\log x=t$, but I can't see that getting anywhere. Please help.


$$I=\int_{0}^{+\infty}\frac{\sqrt{t}}{e^t\sqrt{1+e^{-4t}}}\,dt $$ is a starter, then $$ \frac{1}{\sqrt{1+z}}=\sum_{n\geq 0}\frac{\binom{2n}{n}}{4^n}(-1)^n z^n $$ leads to $$ I = \sum_{n\geq 0}\frac{\binom{2n}{n}}{4^n}(-1)^n \int_{0}^{+\infty}\sqrt{t}\, e^{-(4n+1)t}\,dx $$ and to $$ I = \frac{\sqrt{\pi}}{2}\sum_{n\geq 0}\frac{\binom{2n}{n}(-1)^n}{4^n(4n+1)^{3/2}}.$$ Not a "closed form" expression, but a pretty good series representation for numerical purposes.
The original integral is related to the semiderivative at the origin of $$ f(s)=\int_{1}^{+\infty}\frac{x^s}{\sqrt{x^4+1}}\,dx = \frac{1}{1-s}\,\phantom{}_2 F_1\left(\tfrac{1}{2},\tfrac{1-s}{4};\tfrac{5-s}{4};-1\right).$$

  • 4
    $\begingroup$ are you a wizard? (+1) $\endgroup$ – clathratus Nov 1 '18 at 1:16
  • 7
    $\begingroup$ No, I am Darkwing Duck :D $\endgroup$ – Jack D'Aurizio Nov 1 '18 at 1:17

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