Asymptotics of the logarithmic integral Problem
Given
$$
\gamma = \int_0^1 {1-e^{-u} \over u} du - \int_1^\infty {e^{-u} \over u} du,
$$
prove that
$$
\int_0^x {dt \over \log t} = \gamma + \log \log x + \sum_{k=1}^\infty {\log^k x \over k \cdot k!}.
$$


*

*Hint: Let $u = \log t$.

*Notes: $\gamma$ is the Euler-Mascheroni constant and $\log x$ is the natural logarithm.
Progress
To use the substitution, we need $u = \log t$, $e^u = t$, $du = dt/t$, and $e^u du = dt$. Plugging these into the first integrals gives
$$
\gamma = \int_1^e {1 - 1/t \over \log t} {dt \over t} - \int_e^\infty {1/t \over \log t} {dt \over t} = \int_1^e {t-1 \over t^2 \log t} dt - \int_e^\infty {dt \over t^2 \log t}.
$$
Using the hint on the desired integral gives
$$
\int_{-\infty}^{\log x} {e^u du \over u} = \gamma + \log \log x + \sum_{k=1}^\infty {\log^k x \over k \cdot k!},
$$
and this series sort of looks like
$$
\sum_{k=1}^\infty {(-\log x)^k \over k \cdot k!} = -\int_0^{\log x} {1 - e^{-u} \over u} du,
$$
which I got from expanding the integrand into a power series. This makes me think to split the intervals $[0,1]$ and $[1,\infty)$ up at the point $\log x$. That gave me
$$
\gamma = \int_0^{\log x} {1 - e^{-u} \over u} du - \log \log x - \int_{\log x}^\infty {e^{-u} \over u} du,
$$
or using the value of the series,
$$
\gamma + \log \log x = \sum_{k=1}^\infty {(-\log x)^k \over k \cdot k!} - \int_{\log x}^\infty {e^{-u} \over u} du.
$$
For this integral, using the substitution from the hint and then $s = 1/t$ gives
$$
\int_{\log x}^\infty {e^{-u} \over u} du = \int_x^\infty {dt \over t^2 \log t} = -\int_0^{1/x} {ds \over \log s}.
$$
Replacing $x$ with $1/x$ would give the correct series with the wrong sign, and the correct upper limit for this last integral, but then the $\log \log x$ terms would be ruined. This seems like there might be a sign error somewhere.
 A: Unfortunately I don't see a way to gently nudge you in the right direction from your start, so I'll begin with the beginning.
If you know where you want to end, it's often very helpful to work from the target towards the starting point too. If we do that here, from the target we obtain
$$\begin{align}
\gamma &+ \log \log x + \sum_{k=1}^\infty \frac{\log^k x}{k\cdot k!}\\
&= \int_0^1 \frac{1-e^{-u}}{u}\,du - \int_1^\infty \frac{e^{-u}}{u}\,du + \int_1^{\log x} \frac{du}{u} + \sum_{k=1}^\infty \frac{\log^k x}{k\cdot k!}\\
&= \int_0^{\log x} \frac{1-e^{-u}}{u}\,du - \int_{\log x}^\infty \frac{e^{-u}}{u}\,du + \sum_{k=1}^\infty \frac{\log^k x}{k\cdot k!}\\
&= \int_0^{\log x} \frac{1-e^{-u}}{u}\,du - \int_{\log x}^\infty \frac{e^{-u}}{u}\,du + \sum_{k=1}^\infty \frac{1}{k!} \int_0^{\log x} u^{k-1}\,du\\
&= \int_0^{\log x} \frac{1-e^{-u}}{u}\,du - \int_{\log x}^\infty \frac{e^{-u}}{u}\,du + \int_0^{\log x}\sum_{k=1}^\infty \frac{1}{k!} u^{k-1}\,du\\
&= \int_0^{\log x} \frac{1-e^{-u}}{u}\,du - \int_{\log x}^\infty \frac{e^{-u}}{u}\,du + \int_0^{\log x} \frac{e^u-1}{u}\,du\\
&= \int_0^{\log x} \frac{e^u-e^{-u}}{u}\,du - \int_{\log x}^\infty \frac{e^{-u}}{u}\,du.
\end{align}$$
Now, the integral
$$\int_0^x \frac{dt}{\log t}$$
exists only in the principal value sense, since the singularity at $1$ isn't integrable. Instead of taking out a completely symmetric interval around $1$, let us consider
$$\lim_{\varepsilon \searrow 0} \int_0^{e^{-\varepsilon}} \frac{dt}{\log t} + \int_{e^{\varepsilon}}^x \frac{dt}{\log t}$$
for $x > 1$. It is easy to see that that limit is the same as the one you get by removing the interval $(1-\varepsilon, \, 1+\varepsilon)$.
Now substituting $u = \pm \log t$ produces something from which the meeting point is easily reached.
