After solving this problem I found out that: $$\int_0^1\ln^sxdx=(-1)^s\Gamma(s+1)\tag{1}$$Where $\Gamma(s)$ is the gamma function, defined as $(s-1)!$ when $s\in\mathbb{Z}$. $(1)$ 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?

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    $\begingroup$ On $(0,1)$ the inner logarithm is negative, therefore there must be an issue as the result is real, yet the outer logarithm is taken out of a negative number. // It seems you forgot about the $(-1)^s$ term. $\endgroup$
    – Zacky
    Jan 2, 2023 at 17:56
  • $\begingroup$ @Zacky I just realized :/ don't know what to do now. $\endgroup$ Jan 2, 2023 at 18:01
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    $\begingroup$ @Zacky Thanks! I will edit the OP. I was confused since $s$ was an independent variable but then I remembered that we are integrating with respect to $x$. $\endgroup$ Jan 2, 2023 at 18:06
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    $\begingroup$ For the integral $\int_{0}^{1}\ln\left(-\ln\left(x\right)\right)dx$, you can just let $u=\ln(x)$ to get $\int_{0}^{\infty}e^{-x}\ln\left(x\right)dx$. There are plenty of methods on this site to show that new integral equals $\gamma$. There is a way to compute $\int_{0}^{1}\ln\left(-\ln\left(x\right)\right)dx$ directly using contour integration, but it just leads to $\int_{0}^{\infty}e^{-x}\ln\left(x\right)dx$ and is a lot more tedious and (I guess) roundabout. If you are interested, I can type an answer for that. $\endgroup$ Jan 3, 2023 at 7:47
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    $\begingroup$ On second thought, I found a different way to avoid $\int_{0}^{\infty}e^{-x}\ln(x)dx$. I'm currently drafting out my thoughts, then I'll type everything in an answer, hopefully by tonight. $\endgroup$ Jan 4, 2023 at 3:04

3 Answers 3


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.

enter image description here

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,

$$\int_{0}^{1}\ln\left(\ln\left(\frac{1}{x}\right)\right)dx = -\gamma.$$


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    $\begingroup$ I didn't expect a Complex Analysis proof! But I won't accept this yet since I would like to see other proofs. I did upvote though :) $\endgroup$ Jan 4, 2023 at 17:16
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    $\begingroup$ Thank you! You don't have to accept it if you don't want to. I don't care that much about points. All I care about is getting my ideas out there. @KamalSaleh $\endgroup$ Jan 4, 2023 at 22:31
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    $\begingroup$ So you won't care to get +15 points? ;) I just realized how much effort you and your friend had on not just the derivation but the graph too. $\endgroup$ Jan 5, 2023 at 5:07
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    $\begingroup$ My friend was the one who made the graph on Desmos, and it was for two American Mathematical Monthly problems that involved Apery's Constant. That contour, coincidentally, was also useful for this problem. I don't think she ever knew how the contour worked mathematically, so it was just me making up the proofs. @KamalSaleh $\endgroup$ Jan 5, 2023 at 5:12
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    $\begingroup$ Complex analysis to the rescue as always xd. Very nice $\endgroup$ Jan 5, 2023 at 20:57

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)

  • $\begingroup$ Great answer! The solution is actually done at step 1 since it is known that this integral is $-\gamma$. Also, by some coincidence, a person recommended this book to me yesterday! $\endgroup$ Jan 6, 2023 at 0:15

Edit: This is the same as a deleted answer I posted a while ago, but it had an un-explained downvote so I will post it again.

We could just substitute $e^{-t}=x$. The derivative is $-e^{-t}=-x$ and $\ln t=\ln\ln\frac{1}{x}$. Then everything falls into place.


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