Integrate $\displaystyle\int_0^1(x-x^{1/x})dx$
I started doing something like:
$u=\ln x\longrightarrow e^u=x\longrightarrow dx=x du=e^udu$
So the integral in the $u$-world goes like this:
$$\displaystyle\int_0^1(e^u-\exp{e^{-u}})dx$$
But I have no idea how to follow. All I end up is just garbage. Thanks.
I got the result in Wolfram Mathematica and it is around $0.146503$, but I don't know how to get that value.
The first part is easy of course. It's the second part of the integral that's the headache. It turns out Donald Splutterwit is absolutely right above in his comment of the question, it's solved using John Bernoulli's Sophomore Dream identity and a real head scratching gamma function substitution. It turns out the answer was given in detail at this site several years ago here.
This is the kind of sadistic calculation British mathematicians used to torture their students with on the Cambridge Tripos exam in analysis. Sadly, I suspect it's the kind of question the Math GRE might put on the exam in the calculus section to trip a student trying desperately to get into graduate school at Columbia or Yale.
\begin{eqnarray} \int_0^{\infty} x^{\frac{1}{x}} dx \end{eqnarray} Break the interval into $[0,1]$ and $[1,\infty)$. One half can be done using https://en.wikipedia.org/wiki/Sophomore%27s_dream ... I think this is the tricky half, but there were some nice answers. – Donald Splutterwit Sep 03 '17 at 22:04