$$\int_0^1 \sqrt{-\ln x} dx$$ I'm looking for alternative methods to what I already know (method I have used below) to evaluate this Integral. $$y=-\ln x$$

$$\bbox[8pt, border:1pt solid crimson]{e^y=e^{-\ln x}=e^{\ln\frac{1}{x}}=\frac{1}{x}}$$

$$\color{#008080}{dx= -\frac{dy}{e^y}}$$

$$\int_0^1 \sqrt{-\ln x} dx=\int_{\infty}^0 \sqrt{y} \left(-\frac{dy}{e^y}\right)=\int_0^{\infty} e^{-y} y^{\frac{1}{2}}dy=\left(\frac{1}{2}\right)!$$

$$\bbox[8pt, border:1pt solid crimson]{\left(\frac{1}{2}\right)!=\frac{1}{2} \sqrt{\pi}}$$

$$\Large{\color{crimson}{\int_0^1 \sqrt{-\ln x} dx=\frac{1}{2} \sqrt{\pi}}}$$

  • $\begingroup$ Do you mean different methods completely, or are different substitutions also ok? You could solve it with $-\ln x = y^2$ as well for instance. $\endgroup$
    – Slugger
    Dec 25, 2014 at 12:29
  • $\begingroup$ @Slugger Different Substitutions too :) Il be glad to receive as many answers as possible :) $\endgroup$
    – The Artist
    Dec 25, 2014 at 13:44
  • $\begingroup$ @TheArtist in the integral you should have $y^{+\frac{1}{2}}$. $\endgroup$
    – alexjo
    Dec 25, 2014 at 15:41
  • $\begingroup$ @alexjo oops a typo :) thanks for spotting it $\endgroup$
    – The Artist
    Dec 25, 2014 at 15:46

4 Answers 4


Here's slightly different approach

We have $$\mathcal I=\int_0^1 \sqrt{-\ln x} \mathrm dx=\int_0^1 \left[\ln\frac1 x\right]^{1/2} \mathrm dx$$ By using IBP $$\left[\ln\frac1 x\right]^{1/2}=u\iff\mathrm du= -\frac 1{2x}\left[\ln\frac1 x\right]^{-1/2}\mathrm dx$$

$$\mathrm dx=\mathrm dv\iff x=v$$

$$\begin{align} \mathcal I &=x\left[\ln\frac1 x\right]^{1/2}\Bigg\rvert_0^1+ \frac 1{2}\int_0^1\left[\ln\frac1 x\right]^{-1/2}\mathrm dx\\ &=\frac 1{2}\int_0^1\left[\ln\frac1 x\right]^{-1/2}\mathrm dx\\ \end{align}$$ Now by substututing $$t^2=\ln\frac1x\implies e^{-t^2}=x\iff \mathrm dx=-2te^{-t^2}\mathrm dt$$ We get $$\mathcal I=\int _0^\infty e^{-t^2}\mathrm dt=\frac{\sqrt\pi}{2}$$ Hence,

$$\int_0^1 \sqrt{-\ln x} \mathrm dx=\frac{\sqrt\pi}{2}$$


As far as I know, there are essentially two ways for proving that $\Gamma\left(\frac{1}{2}\right)=\sqrt{\pi}$.

The first way is to use integration by parts, leading to $\Gamma(z+1)=z\Gamma(z)$, in order to relate $\Gamma\left(\frac{1}{2}\right)$ to the Wallis product. The latter can be computed by exploiting the Weierstrass product for the sine function: $$\frac{\sin z}{z}=\prod_{n=1}^{+\infty}\left(1-\frac{z^2}{\pi^2n^2}\right)$$ by evaluating it in $z=\frac{\pi}{2}$. The duplication formula for the $\Gamma$ function follows from this approach.

The second way is to use some substitutions in order to relate $\Gamma\left(\frac{1}{2}\right)$ to the gaussian integral $$\int_{-\infty}^{+\infty}e^{-x^2}\,dx$$ that can be evaluated through Fubini's theorem and polar coordinates.

  • 1
    $\begingroup$ (+1) Oh I see, I used the first way. Thank you very much :) $\endgroup$
    – The Artist
    Dec 25, 2014 at 15:50

Similar to what Iuʇǝƃɹɐʇoɹ did, let $u=\sqrt{-\ln x}$. Then $x=e^{-u^2}$ and hence $$ \int_0^1\sqrt{-\ln x}dx=-\int_0^\infty ud(e^{-u^2})=-ue^{-u^2}|_0^\infty+\int_0^\infty e^{-u^2}du=\frac{\sqrt{\pi}}{2}. $$


Here are two more variations of the theme. According to OPs presentation and the characterisation of @JackDAurizio the essential point is to show the identity


The first variant is based upon the beta integral in combination with a substitution to obtain the identity.

The second one is much more exciting. It's not the calculation itself which is routine work, but the new concept based upon periods introduced at the end of the last millennium.

1. Variant: Beta integral with substitution

Following Special functions by G.E. Andrews, R. Askey and R. Roy we recall the definition of the beta-integral

\begin{align*} B(x,y)=\int_0^{1}t^{x-1}(1-t)^{y-1}dt\qquad\qquad \operatorname{Re} x>0,\operatorname{Re} y>0 \end{align*}

and the well known identity

\begin{align*} B(x,y)=\frac{\Gamma(x)\Gamma(y)}{\Gamma(x+y)} \end{align*}

This symmetry of the beta integral gives rise to the substitution $t=\sin^2\theta$. We obtain

\begin{align*} B(x,y)=2\int_0^{\frac{\pi}{2}}\sin^{2x-1}\theta\cos^{2y-1}\theta d\theta=\frac{\Gamma(x)\Gamma(y)}{\Gamma(x+y)} \end{align*}

Put $x=\frac{1}{2},y=\frac{1}{2}$, to get

\begin{align*} B\left(\frac{1}{2},\frac{1}{2}\right)=2\int_0^{\frac{\pi}{2}}d\theta=\pi=\frac{\Gamma(\frac{1}{2})^2}{\Gamma(1)} \end{align*}

and $\Gamma\left(\frac{1}{2}\right)=\sqrt{\pi}$ follows.

Another substitution of the beta integral will be convenient in the next variant.

Variant 2: $\Gamma\left(\frac{1}{2}\right)$ and $\sqrt{\pi}$ are periods forming an accessible identity

The definition of periods below and the proof is based upon the fascinating introductory survey paper about periods by M. Kontsevich and D. Zagier.

Periods are defined as complex numbers whose real and imaginary parts are values of absolutely convergent integrals of rational functions with rational coefficient over domains in $\mathbb{R}^n$ given by polynomial inequalities with rational coefficients.

The set of periods is therefore a countable subset of the complex numbers. It contains the algebraic numbers, but also many of famous transcendental constants. One can also replace rational functions with algebraic functions and rational coefficient with algebraic coefficient without changing the set of numbers $\mathcal{P}$.

In order to show the identity $\Gamma\left(\frac{1}{2}\right)=\sqrt{\pi}$ we first have to show according to Kontsevich's and Zagiers paper that both $\Gamma\left(\frac{1}{2}\right)$ and $\sqrt{\pi}$ are periods and that they form a so-called accessible identity.

First step of the proof: $\Gamma\left(\frac{1}{2}\right)$ and $\pi$ are periods

There are a lot of different proper representations of $\pi$ showing that this constant is a period. In the referred paper above following expressions (besides others) of $\pi$ are stated:

\begin{align*} \pi= \iint \limits_{x^2+y^2\leq 1}dxdy=\int_{-\infty}^{\infty}\frac{dt}{1+t^2}=\int_{-1}^{1}\frac{dt}{\sqrt{1-t^2}}\tag{1} \end{align*}

showing that $\pi$ is a period. According to the paper values of the gamma function $$\Gamma(s)=\int_{0}^{\infty}t^{s-1}e^{-t}dt$$ at rational values of the argument $s$ are closely related to periods: $$\Gamma(p/q)^q\in \mathcal{P}\qquad\qquad (p,q\in\mathbf{N}).$$ This follows from the repesentation of $\Gamma(p/q)^q$ as beta integral. Putting $p=1$ and $q=2$ it follows that $\Gamma(1/2)^2$ is a period.

We conclude: $\Gamma\left(\frac{1}{2}\right)$ is a period.

$$ $$

Second step: $\Gamma\left(\frac{1}{2}\right)$ and $\sqrt{\pi}$ form an accessible identity.

An accessible identity between two periods $A$ and $B$ is given, if we can transform the integral representation of period $A$ by application of the three rules: Additivity (integrand and domain), Change of variables and Newton-Leibniz formula to the integral representation of period $B$.

This implies equality of the periods and the job is done.

We again consider the beta integral and take the substitution $t=\frac{u-a}{b-a}$. We get \begin{align*} \int_a^b(b-u)^{x-1}(u-a)^{y-1}du &=(b-a)^{x+y-1}B(x,y)\\ &=(b-a)^{x+y-1}\frac{\Gamma(x)\Gamma(y)}{\Gamma(x+y)} \end{align*}

The special case $a=-1, b=1$ gives

\begin{align*} \int_{-1}^{1}(1+t)^{x-1}(1-t)^{y-1}dt=2^{x+y-1}\frac{\Gamma(x)\Gamma(y)}{\Gamma(x+y)} \end{align*}

Put $x=y=\frac{1}{2}$ to get

\begin{align*} \int_{-1}^{1}\frac{1}{\sqrt{1-t^2}}dt=\Gamma\left(\frac{1}{2}\right)^2=\pi\tag{2} \end{align*}

The identities in (1) and (2) show that $\Gamma\left(\frac{1}{2}\right)$ and $\sqrt{\pi}$ form an accessible identity and the claim $\Gamma\left(\frac{1}{2}\right)=\sqrt{\pi}$ follows.


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