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Consider the integral $$I=\int_0^\infty \frac{f(x)}{1+x^2}\mathrm{d}x$$ now make a change of variable to $x=\dfrac{1}{t}$, and so $\mathrm{d}x=-\dfrac{1}{t^2}\mathrm{d}t$. The integral becomes $$I=\int_0^\infty \frac{f\left(\frac{1}{t}\right)}{1+t^2}\mathrm{d}t$$ and so $$I=\frac{1}{2}\int_0^\infty \frac{f(x)+f\left(\frac{1}{x}\right)}{1+x^2}\mathrm{d}x$$ Hence the question: what are the functions $f(x)$ such that the quantity $L(f)=f(x)+f\left(\frac{1}{x}\right)$ has a "nice" value that makes the integral easy?

For example, some uninteresting cases are:

  1. $L(\ln(x))=0$ that provides $\displaystyle\int_0^\infty \dfrac{\ln(x)}{1+x^2}\mathrm{d}x=0$

  2. $L(\arctan(x))=\frac{\pi}{2}$ that provides $\displaystyle\int_0^\infty \dfrac{\arctan(x)}{1+x^2}\mathrm{d}x=\dfrac{\pi^2}{8}$

  3. $L(\sin(\ln(x)))=0$ then $\displaystyle\int_0^\infty \dfrac{\sin(\ln(x))}{1+x^2}\mathrm{d}x=0$

and in general every time $g(x)$ is odd and the integral still converges, then $\displaystyle\int_0^\infty \dfrac{g(\ln(x))}{1+x^2}\mathrm{d}x=0$

I wonder if there are some other special or lesser known functions that have the charateric of having a nice $L(f)$. Tell me if you come up with something.

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Zima
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    There is even a book on a similar subject. P. Koosis, "The logarithmic integral", Cambridge university press. – Giuseppe Negro Apr 26 '23 at 14:43
  • $\ln$ and $\arctan$ are special in that they satisfy identities relating an argument $x$ to its reciprocal $\frac1x$, namely $\ln\left(\frac1x\right)=-\ln(x)$ and $\arctan\left(\frac1x\right)=\frac\pi2-\arctan(x)$ (provided $x>0$). I think (but cannot prove) that a "good" $f$ must possess this sort of property. – user170231 Apr 26 '23 at 17:24
  • Yes that is what I mean with "nice value for $L(f)$" – Zima Apr 26 '23 at 17:25
  • Related. – TheSilverDoe Apr 26 '23 at 19:48

1 Answers1

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A slightly more difficult example:

The inversion formula for the dilogarithm says that $$\operatorname{Li}_{2}(x) + \operatorname{Li}_{2} \left( \frac{1}{x} \right) = - \zeta(2) - \frac{1}{2} \, \ln^{2} (-x)$$ for $x \notin [0, 1)$.

Therefore, we have $$ \begin{align} \int_{0}^{\infty} \frac{\operatorname{Li}_{2}(-x)}{1+x^{2}} \, \mathrm dx &= \frac{1}{2} \int_{0}^{\infty} \frac{\operatorname{Li}_{2}(-x) + \operatorname{Li}_{2} \left(- \frac{1}{x} \right) }{1+x^{2}} \, \mathrm dx \\ &= - \frac{1}{2} \int_{0}^{\infty} \frac{\zeta(2) + \frac{1}{2} \, \ln^{2}(x)}{1+x^{2}} \, \mathrm dx \\ &\overset{(1)}{=} - \frac{1}{2} \left(\zeta(2) \, \frac{\pi}{2} + \frac{\pi^{3}}{16} \right) \\ &= - \frac{1}{2} \left(\frac{\pi^{3}}{12} + \frac{\pi^{3}}{16}\right) \\ &= - \frac{7 \pi^{3}}{96}. \end{align}$$

$(1)$ Prove $\int_0^\infty \frac{\log^2(x)}{x^2+1}\mathrm dx=\frac{\pi^3}{8}$ with real methods

Help with integrating $\int_0^{\infty} \frac{(\log x)^2}{x^2 + 1} \operatorname d\!x$ - contour integration?

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