In the Knockout Match 3 in the MIT integration bee 2022 participants were asked to compute the following integral:
\begin{align}
\int_0^{+\infty}{\sum_{k = \lfloor{x}\rfloor}^{+\infty}{\frac{2^{-k}}{k+1}}\mathrm{d}x}\tag{1} \end{align}
and the given answer is $4$. However, when I tried to compute the integral, I did not get $4$ and perhaps my logic is somehow flawed. First, I tried to break the integral into an infinite sum of definite integrals so $(1)$ becomes:
\begin{align}
\int_0^{+\infty}{\sum_{k = \lfloor{x}\rfloor}^{+\infty}{\frac{2^{-k}}{k+1}}\mathrm{d}x}&=\lim_{t \to +\infty}{\sum_{n = 0}^{t}\left[{\int_{n}^{n+1}{\left(\sum_{k= \lfloor{x}\rfloor}^{+\infty}{\frac{2^{-k}}{k+1}}\right)\mathrm{d}x}}\right]} \\
&=\lim_{t \to +\infty}{\sum_{n=0}^{t}\left[{\int_{n}^{n+1}{\left(\sum_{k = n}^{+\infty}{\frac{2^{-k}}{k+1}}\right)\mathrm{d}x}}\right]} \\
&= \lim_{t \to +\infty}{\sum_{n=0}^{t}{\left[2^{-n}\int_{n}^{n+1}{\left(\sum_{k=0}^{+\infty}{\frac{2^{-k}}{k+n+1}}\right)\mathrm{d}x}\right]}} \\
&= \lim_{t \to +\infty}{\sum_{n=0}^{t}}{\left[2^{-n}\Phi\left(\frac{1}{2}, \ 1,\ n+1\right)\right]} \tag{2}\\
\end{align}
where $\Phi$ is the Lerch transcendent function. But we have that
\begin{align}
\Phi\left(\frac{1}{2},\ 1,\ n+1\right) &= \int_{0}^{\infty}{\frac{e^{-(n+1)t}}{1-\frac{1}{2}e^{-t}}dt} \\
&= \int_{1}^{\infty}{\frac{1}{\left(1-\frac{1}{2u}\right)u^{n+2}}\mathrm{d}u} \\
&= \int_{1}^{\infty}\left({\frac{2^{n+2}}{2u-1}-\sum_{k=1}^{n+1}{\frac{2^{n-k+2}}{u^{k}}}}\right)\mathrm{d}u \\
&=\Biggr[2^{n+1}\ln{(2u-1)}-2^{n+1}\ln{u}-\sum_{k=2}^{n+1}{2^{n-k+2}\frac{u^{1-k}}{1-k}}\Biggr]_1^{\infty} \\
&= 2^{n+1}\ln{2}-\sum_{k=2}^{n+1}{\frac{2^{n-k+2}}{1-k}}
\end{align}
so $(2)$ becomes:
\begin{align}
\int_0^{+\infty}{\sum_{k = \lfloor{x}\rfloor}^{+\infty}{\frac{2^{-k}}{k+1}}\mathrm{d}x}&=\lim_{t \to +\infty}{\sum_{n=0}^{t}\left({2\ln{2}-2^{-n}\sum_{k=2}^{n+1}}\frac{2^{n-k+2}}{1-k}\right)} \\
&=\lim_{t \to +\infty}{\sum_{n=0}^{t}\left({2\ln{2}-\sum_{k=2}^{n+1}}\frac{2^{-k+2}}{1-k}\right)}
\end{align}
but I can't compute the last sum in terms of $n$ so that I may compute the final infinite sum. I know this is a bit too far gone, but is there maybe some way to save this approach and go forward? Also, any clever solutions of the problem, more elementary, are welcome.
Thanks in advance!
Best Answer
$$ I:=\int_0^\infty\sum_{k=\lfloor x\rfloor}^\infty\frac{2^{-k}}{k+1}\,dx=\sum_{n=0}^\infty{\int_n^{n+1}}\sum_{k=n}^\infty\frac{2^{-k}}{k+1}\,dx =\sum_{n=0}^\infty\sum_{k=n}^\infty\frac{2^{-k}}{k+1}. \tag{1} $$ Writing each internal sum of $(1)$ in a separate line, we obtain \begin{align} I=&\,\, 1+\frac{2^{-1}}{2}+\frac{2^{-2}}{3}+\ldots \\ &\phantom{\,\,1}+\frac{2^{-1}}{2}+\frac{2^{-2}}{3}+\ldots \\ &\phantom{\,\,1+\frac{2^{-1}}{2}}+\frac{2^{-2}}{3}+\ldots \\ &\phantom{\,\,1+\frac{2^{-1}}{2}+\frac{2^{-2}}{3}}+\ldots \tag{2} \end{align} If we now reorder the terms along the vertical lines, we can rewrite $(2)$ as $$ I=1+2\,\frac{2^{-1}}{2}+3\,\frac{2^{-2}}{3}+\ldots=\sum_{n=0}^{\infty}2^{-n}=2, \tag{3} $$ which is the same result obtained by @user170231.