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sujeethan.B recently discovered a generalized form of the following integral:

$$ I(a) = \int_{0}^{\infty} \frac{x \log(1 + e^{-a\pi x})}{1 + x^2} \,dx $$

and found that it can be expressed as

$$ I(a) = \frac{a\pi^2}{4} + \ln \left(\frac{2}{a}\right) \ln \left(2 \cos \left( \frac{a\pi}{2} \right) \right) + \frac{1}{2} \zeta''\left(0, \frac{1}{2} + \frac{a}{2} \right) + \frac{1}{2} \zeta''\left(0, \frac{1}{2} - \frac{a}{2} \right) $$

$$ + \sum_{k=0}^{\infty} \left[ \operatorname{Li}_2 \left( \frac{a}{(2k+1) + a} \right) + \operatorname{Li}_2 \left( \frac{-a}{(2k+1) - a} \right) \right], \quad 0 < a < 1. $$

Question:

  1. Is there a more simplified or alternative form of this result?

  2. Are there any special cases of ( a ) where this result simplifies further?

I would greatly appreciate any insights or references that could help refine this result.

Thank you!

Rishani Rajes
  • 51

1 Answers1

4

Making $\color{red}{a \pi \to a}$ $$I(a) = \int_{0}^{\infty} \frac{x \log(1 + e^{-ax})}{1 + x^2} \,dx$$

Expand the logarithm to face $$\frac{x \log(1 + e^{-ax})}{1 + x^2}=\sum_{n=1}^\infty\frac{ (-1)^{n+1}}n\,\frac x{1+x^2}\,e^{-n a x}$$ and $$\int \frac x{1+x^2}\,e^{-n a x}\,dx=\frac 12\Big(e^{-i a n}\, \text{Ei}(-a n (x-i))+e^{i a n}\, \text{Ei}(-a n (x+i)) \Big)$$ $$J_n(a)=\int_0^\infty \frac x{1+x^2}\,e^{-n a x}\,dx$$ $$\color{blue}{J_n(a)=\frac{\pi}{2} \sin (a n)-(\text{Ci}(a n) \cos (a n)+\text{Si}(a n) \sin (a n))}$$ $$ \left|\frac{J_{n+1}(a)}{J_{n}(a)}\right|\sim 1-\frac 2 n$$ $$\large\color{red}{I(a)=\sum_{n=1}^\infty\frac{ (-1)^{n+1}}n\,J_n(a)}$$ Interesting is $$J_n(\pi)=(-1)^{n+1} \text{Ci}( n\pi)\quad \implies \quad I(\pi)=\sum_{n=1}^\infty \frac{\text{Ci}(n \pi )}{n}$$ which converges reasonably fast. Computing the partial sums $$\left( \begin{array}{cc} p & \sum_{n=1}^p \frac{\text{Ci}(n \pi )}{n}\\ 10 & 0.0648622 \\ 20 & 0.0648993 \\ 30 & 0.0649034 \\ 40 & 0.0649044 \\ 50 & 0.0649048 \\ \end{array} \right)$$

Edit

Limiting the summation to $100$ terms, the results $$\left( \begin{array}{ccc} a & \text{summation} & \text{integration} \\ 0.1 & 1.42893720 & 1.42898400 \\ 0.2 & 1.01442770 & 1.01444093 \\ 0.3 & 0.79394618 & 0.79395191 \\ 0.4 & 0.65074171 & 0.65074532 \\ 0.5 & 0.54857755 & 0.54857953 \\ 0.6 & 0.47150894 & 0.47151087 \\ 0.7 & 0.41115225 & 0.41115484 \\ 0.8 & 0.36258426 & 0.36258581 \\ 0.9 & 0.32268593 & 0.32268671 \\ 1 & 0.28936850 & 0.28936894 \\ 2 & 0.12547474 & 0.12547486 \\ 3 & 0.06970638 & 0.06970643 \\ 4 & 0.04403759 & 0.04403763 \\ 5 & 0.03018394 & 0.03018397 \\ 6 & 0.02190055 & 0.02190056 \\ 7 & 0.01657494 & 0.01657495 \\ 8 & 0.01295922 & 0.01295923 \\ 9 & 0.01039770 & 0.01039771 \\ 10 & 0.00851988 & 0.00851995 \\ \end{array} \right)$$

Claude Leibovici
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