Upper bound of integral $\int_0^b\frac{1}{x+2^x} dx$

Question

Is it known a closed form of the integral: $$\displaystyle\int_0^b\dfrac{1}{x+2^x} dx?$$ Using the Talenti inequality I found the following bound: $$\displaystyle\int_0^b\dfrac{1}{x+2^x} dx\lt\ln\left(1-\dfrac{-1+e^{-b}}{\ln(2)}\right)$$ For $b\gt\gt1$ the difference between the integral and the logarithm in the right doesn't exceed 0.19 Is it known some better result? Obviously it is possible to evaluate numerically this integral.

Answer 1

I believe that you can use a simple Riemann-Integral to determine this upper bound. \begin{eqnarray} \int_0^b \frac{1}{x'+2^{x'}} dx' =& \int_0^1 \frac{1}{x+ b\, 2^{\frac{x}{b}}} dx\\ \int_0^1 \frac{1}{x+b\,2^{\frac{x}{b}}} dx <& \sum_{i=0}^N \frac{1}{N}\frac{1}{\frac{i}{N}+b\,2^{\frac{i}{Nb}}} = \sum_{i=0}^N \frac{1}{i+Nb\, 2^{\frac{i}{Nb}}} \end{eqnarray}