Proof of $\int_0^1\ln\left(\frac{1}{\ln \frac{1}{x}}\right)dx=\gamma$ After solving this problem I found out that: $$\int_0^1\ln^sxdx=(-1)^ss!$$Which can be proven by induction. Anyways, I realized that I could use this to prove that $$-\int_0^1\ln\left(\ln \frac{1}{x}\right)dx=\gamma$$Where $\gamma$ is the Euler-mascheroni constant. Here is my proof:

Take the derivative of $\int_0^1(-\ln x)^sdx$ with respect to $s$. To do this, move the differentiation operator inside the integral sign, meaning that we need to take the derivative of $\ln^sx$. This is the same as $e^{s\ln\ln \frac{1}{x}}$. By chain rule we get that the derivative of $(-\ln x)^s$ is $\ln(-\ln x)(-\ln x)^s$. Setting $s=0$ back in the original equation, we get that $\int_0^1\ln\frac{1}{\ln \frac{1}{x}}dx=-\gamma$ since the derivative of the analytic continuation of the factorial function at zero is $-\gamma$. Multiplying both sides by $-1$ yields the desired expression.

Are there other proofs to this expression?
 A: We will prove
$$\int_{0}^{1}\ln\left(\ln\left(\frac{1}{x}\right)\right)dx = -\gamma.$$
Proof. Let $f(z) = \ln\left(\ln\left(\dfrac{1}{z}\right)\right)$ where $\operatorname{arg}(z) \in (-\pi, \pi]$. Define a contour $C = [\epsilon,1-\epsilon] \cup \gamma_2 \cup \Gamma \cup [i,i\epsilon] \cup \gamma_1$ for small $\epsilon > 0$ that we will traverse counterclockwise around. A visual is provided below, and its Desmos code can be viewed here made by one of my friends.

By Cauchy's Residue Theorem, we write $\displaystyle\oint_C f(z)dz$ as
$$
\eqalign{
0 &= \int_{\epsilon}^{1-\epsilon}f(z)dz + \int_{\gamma_2} f(z)dz + \int_{\Gamma} f(z)dz +\int_{i}^{i\epsilon} f(z)dz + \int_{\gamma_1} f(z)dz \cr
\operatorname{P.V.}\int_0^1 f(z)dz &= \lim_{\epsilon \to 0}\left(\int_{-\gamma_1} f(z)dz + \int_{-\gamma_2} f(z)dz + \int_{i\epsilon}^{i} f(z)dz - \int_{\Gamma} f(z)dz\right).
}
$$

Both $\displaystyle\int_{-\gamma_1} f(z)dz$ and $\displaystyle\int_{-\gamma_2} f(z)dz$ approach $0$ as $\epsilon \to 0$ using similar methods. To demonstrate,
$$
\eqalign{
\lim_{\epsilon \to 0}\int_{-\gamma_1}f(z)dz &= \lim_{\epsilon \to 0}\int_{0}^{\pi/2}f\left(\epsilon e^{i\theta}d\epsilon e^{i\theta}\right) \cr
&= i\int_{0}^{\pi/2}e^{i\theta}\lim_{\epsilon \to 0}\epsilon \ln\left(\ln\left(\frac{1}{\epsilon e^{i\theta}}\right)\right)d\theta \cr
&= i\int_{0}^{\pi/2}e^{i\theta}\lim_{\epsilon \to 0} \frac{\ln\left(-\ln\left(\epsilon e^{i\theta}\right)\right)}{1/\epsilon}d\theta \cr
&= i\int_{0}^{\pi/2}e^{i\theta}\lim_{\epsilon \to 0} \frac{\frac{d}{d\epsilon}\ln\left(-\ln\left(\epsilon e^{i\theta}\right)\right)}{\frac{d}{d\epsilon}1/\epsilon}d\theta \cr
&= i\int_{0}^{\pi/2}e^{i\theta}(0)d\theta \cr
&= 0.
}
$$
We can apply a similar procedure for $\displaystyle\int_{-\gamma_2} f(z)dz$, albeit a bit more grunt work. We would have to parameterize $z=1+\epsilon e^{i\theta}$ for $\theta \in [\pi/2, \pi]$.

For $\displaystyle \int_{\Gamma} f(z)dz$, let $z = e^{it}$ such that $t \in [t_{\epsilon}, \pi/2]$. Note that $t_{\epsilon} > 0$ is small and accounts for the tiny gap near $z=1$. Then
$$
\eqalign{
\int_{t_{\epsilon}}^{\pi/2}f\left(e^{it}\right)de^{it} &= i\int_{t_{\epsilon}}^{\pi/2} \ln\left(-\ln\left(e^{it}\right)\right)e^{it}dt \cr
&= i\int_{t_{\epsilon}}^{\pi/2}\ln\left(-it\right)e^{it}dt \cr
&= i\int_{t_{\epsilon}}^{\pi/2} e^{it}\left(\ln\left|-it\right|+i\operatorname{arg}\left(-it\right)\right)dt \cr
&= i\int_{t_{\epsilon}}^{\pi/2} e^{it}\left(\ln\left(t\right)-\frac{i\pi}{2}\right)dt \cr
&= i\Big[i\operatorname{Ei}(it)-ie^{it}\ln(t)\Big]_{t_{\epsilon}}^{\pi/2} + \frac{\pi}{2}+\frac{i\pi}{2} \cr
&\to \gamma + \frac{\pi}{2}+i\pi+i\ln\left(\frac{\pi}{2}\right)- \operatorname{li}(i)
}
$$
as $t_{\epsilon} \to 0$, where we used the definitions of the Exponential Integral and the Logarithmic Integral.

We evaluate $\displaystyle \int_{i\epsilon}^{i} f(z)dz$ as follows:
$$
\eqalign{
\int_{i\epsilon}^{i} \ln\left(-\ln\left(z\right)\right)dz &= \int_{\epsilon}^{1} \ln\left(-\ln\left(iy\right)\right)diy \cr
&= i\Big[y\ln\left(-\ln\left(iy\right)\right)\Big]_{\epsilon}^{1} - i\int_{\epsilon}^{1}\frac{dy}{\ln\left(iy\right)} \cr
&= i\left(\ln\left(-\ln\left(i\right)\right)-\epsilon\ln\left(-\ln\left(i\epsilon\right)\right)\right) - i\Big[-i\operatorname{li}(iy)\Big]_{\epsilon}^{1} \cr
&\to \frac{\pi}{2}+i\pi+i\ln\left(\frac{\pi}{2}\right)- \operatorname{li}(i)
}
$$
as $\epsilon \to 0$.

Going back to $\displaystyle \oint_C f(z)dz$, we collect our results:
$$
\eqalign{
\operatorname{P.V.}\int_0^1 f(z)dz &= 0 + 0 + \left(\frac{\pi}{2}+i\pi+i\ln\left(\frac{\pi}{2}\right)- \operatorname{li}(i)\right) - \left(\gamma + \frac{\pi}{2}+i\pi+i\ln\left(\frac{\pi}{2}\right)- \operatorname{li}(i)\right) \cr
&= -\gamma. \cr
}
$$
In conclusion,
$$\int_{0}^{1}\ln\left(\ln\left(\frac{1}{x}\right)\right)dx = -\gamma.$$
Q.E.D.
A: You can find the answer in Nahin's book "Inside Interesting Integrals". It contains a great number of amazing integration techniques. You absolutely can't miss the book if you are fond of definite integrals
I may briefly illustrate the solution.
Step1. set $y = \ln{\frac{1}{x}}$ and the integral is reduced to $\int_0^{+\infty}e^{-y}\ln{y}\;\mathrm{d} y$
Step2. $\int_0^{+\infty} e^{-x}\ln{x}\;\mathrm{d}x = \int_0^{1} e^{-x}\ln{x}\;\mathrm{d}x + \int_1^{+\infty} e^{-x}\ln{x}\;\mathrm{d}x$
Step3. notice that $e^{-x} = -\frac{\mathrm{d}}{\mathrm{d}x}(e^{-x}-1)$
then we integrate by parts : $\int_0^{1} e^{-x}\ln{x}\;\mathrm{d}x = -\int_0^1\ln{x} \frac{\mathrm{d}}{\mathrm{d}x}(e^{-x}-1)\;\mathrm{d}x = -[\ln{x}(e^{-x}-1)]\big|_0^1 - \int_0^1 \frac{1-e^{-x}}{x}\;\mathrm{d}x = - \int_0^1 \frac{1-e^{-x}}{x}\;\mathrm{d}x$
Step4. $\int_1^{+\infty}e^{-x}\ln{x}\;\mathrm{d}x = \int_1^{+\infty} \frac{e^{-x}}{x}\;\mathrm{d}x$
Step5. Put them together : $\int_0^{+\infty} e^{-x}\ln{x}\;\mathrm{d}x = - (\int_0^1 \frac{1-e^{-x}}{x}\;\mathrm{d}x - \int_1^{+\infty} \frac{e^{-x}}{x}\;\mathrm{d}x)$
Nahin proved in his book that $\int_0^1 \frac{1-e^{-x}}{x}\;\mathrm{d}x - \int_1^{+\infty} \frac{e^{-x}}{x}\;\mathrm{d}x = \gamma$
first, he considered the following integral:$\int_0^1 \frac{1-(1-x)^n}{x}\;\mathrm{d}x = H(n) = \sum_{i=0}^n \frac{1}{i}$
then, he set $u = nx$ and wrote : $H(n) = \int_0^1 \frac{1-(1-\frac{u}{n})^n}{u}\;\mathrm{d}u + \ln{n} - \int_1^n \frac{(1-\frac{u}{n})^n}{u}\;\mathrm{d}u$
let $n \to \infty$ : $\int_0^1 \frac{1-e^{-x}}{x}\;\mathrm{d}x - \int_1^{+\infty} \frac{e^{-x}}{x}\;\mathrm{d}x = \gamma$ (this step is not rigour, you may complete it with more proofs)
