Is possible to use "Feynman's trick" (differentiate under the integral or Leibniz integral rule) to calculate $\int_0^1 \frac{\ln(1-x)}{x}dx\:?$

Question

I heard that the equivalent integral: $-\int_0^\infty \frac{x}{e^x-1}dx$ can be done using Contour integration (I never studied this). Also that sometimes "Leibniz integral rule" is used instead of Contour integration. So can "Feynman's trick" be used to show that $\int_0^1 \frac{\ln(1-x)}{x}dx = -\frac{\pi^2}{6}$ $\:\:?$

Answer 1

A direct application might be

$$\left.\frac{d}{ds} \int_0^{1-\delta} \frac{x^s}{1-x} \, dx\right|_{s = 0} = \left.\int_0^{1-\delta} \frac{ x^s \ln x}{1-x} \, dx\right|_{s = 0} = \int_0^{1-\delta} \frac{\ln x}{1-x} \, dx = \int_\delta^{1 } \frac{\ln (1-x)}{x} \, dx $$

Evaluate the integral on the LHS using the geometric series expansion expansion of $1/(1-x)$ and then take the limit as $\delta \to 0$ (since the improper integral on the RHS converges).

You can also proceed by proving the second equality in

$$- \int_0^1 \frac{\ln(1-x)}{x} \,dx = \int_0^1 \int_0^1 \frac{1}{1 - xy}\, dx \,dy = \sum_{k=1}^\infty \frac{1}{k^2} =\zeta(2) = \frac{\pi^2}{6}$$

using the geometric series $1/(1 -xy) = 1 + xy + (xy)^2 + \ldots $ The first equality is fairly obvious.

More generally we get by the same process

$$\int_0^1 \int_0^1 \frac{x^\alpha y^\alpha}{1 - xy}\, dx \,dy = \sum_{k=1}^\infty \frac{1}{(k + \alpha)^2}$$

and Feynman's trick of repeated integration with respect to $\alpha$ is used to extend the result to other integrals.

Answer 2

Let $\displaystyle J=\int_0^1 \frac{\ln(1-x)}{x}\,dx$

Let $f$ be a function defined on $\left[0;1\right]$,

$\displaystyle f(s)=\int_0^{\frac{\pi}{2}} \arctan\left(\frac{\cos t-s}{\sin t}\right)\,dt$

Observe that,

$\begin{align} f(0)&=\int_0^{\frac{\pi}{2}}\arctan\left(\frac{\cos t}{\sin t}\right)\,dt\\ &=\int_0^{\frac{\pi}{2}} \left(\frac{\pi}{2}-t\right)\,dt\\ &=\left[\frac{t(\pi-t)}{2}\right]_0^{\frac{\pi}{2}}\\ &=\frac{\pi^2}{8} \end{align}$

$\begin{align} f(1)&=\int_0^{\frac{\pi}{2}}\arctan\left(\frac{\cos t-1}{\sin t}\right)\,dt\\ &=\int_0^{\frac{\pi}{2}}\arctan\left(-\tan\left(\frac{t}{2}\right)\right)\,dt\\ &=-\int_0^{\frac{\pi}{2}}\arctan\left(\tan\left(\frac{t}{2}\right)\right)\,dt\\ &=-\int_0^{\frac{\pi}{2}} \frac{t}{2}\,dt\\ &=-\frac{\pi^2}{16} \end{align}$

For $0<s<1$,

$\begin{align} f^\prime(s)&=-\int_0^{\frac{\pi}{2}}\frac{\sin t}{1-2s\cos t+s^2}\,dt\\ &=-\Big[\frac{\ln(1-2s\cos t+s^2)}{2s}\Big]_0^{\frac{\pi}{2}}\\ &=\frac{\ln(\left(1-s)^2\right)}{2s}-\frac{\ln(1+s^2)}{2s}\\ &=\frac{\ln(1-s)}{s}-\frac{\ln(1+s^2)}{2s}\\ \end{align}$

Therefore,

$\begin{align} f(1)-f(0)&=\int_0^1 f^\prime(s)\,ds\\ &=\int_0^1 \left(\frac{\ln(1-s)}{s}-\frac{\ln(1+s^2)}{2s}\right)\,ds\\ -\frac{\pi^2}{16}-\frac{\pi^2}{8}&=J-\int_0^1 \frac{\ln(1+s^2)}{2s}\,ds\\ -\frac{3\pi^2}{16}&=J-\int_0^1 \frac{\ln(1+s^2)}{2s}\,ds\\ \end{align}$

In the latter integral perform the change of variable $y=s^2$,

$\begin{align} -\frac{3\pi^2}{16}&=J-\frac{1}{4}\int_0^1 \frac{\ln(1+y)}{y}\,dy\\ &=J-\frac{1}{4}\int_0^1 \frac{\ln(1-y^2)-\ln(1-y)}{y}\,dy\\ &=J+\frac{1}{4}J-\frac{1}{4}\int_0^1 \frac{\ln(1-y^2)}{y}\,dy\\ \end{align}$

In the latter integral perform the change of variable $x=y^2$,

$\begin{align} -\frac{3\pi^2}{16}&=J+\frac{1}{4}J-\frac{1}{4}\times \frac{1}{2}J\\ &=\frac{9}{8}J\\ \end{align}$

Therefore,

$\begin{align}J&=\frac{8}{9}\times -\frac{3}{16}\pi^2\\ &=\boxed{-\frac{\pi^2}{6}}\end{align}$

Answer 3

I assume you're not happy with $$ \ln(1-x) = -\sum_{n=1}^\infty \frac{x^n}{n}, \qquad x\in(-1,1) $$ from which $$\begin{align} \int_0^1 \frac{\ln(1-x)}{x}dx &= -\int_0^1 \sum_{n=1}^\infty \frac{x^{n-1}}{n} dx = -\int_0^1 \sum_{n=0}^\infty \frac{x^{n}}{n+1} dx \\&\stackrel{\rm (\ast)}{=} -\sum_{n=0}^\infty \frac{1}{n+1}\int_0^1 x^n dx = -\sum_{n=0}^\infty \frac{1}{(n+1)^2}\\ &= -\sum_{n=1}^\infty \frac{1}{n^2} = \boxed{-\frac{\pi^2}{6}} \end{align}$$ ? (It's not Feynman's trick, just a nice series representation for $\ln(1-x)$ which goes a long way.)

The only "catch" here is that swapping $\int$ and $\sum$ in $(\ast)$ actually requires a little bit of justification.

Answer 4

You can use the beta function to do that, namely $$B(x,y) = \int _0^{1} t^{x-1} (1-t)^{y-1 } \mathop{\text{d} x}$$ We will use that $$ \frac{\partial B(x,y)}{\partial y}=B(x,y)( \psi(y) - \psi(x+y))$$ where $\psi$ is the digamma function.

Therefore, $$\frac{\partial B(x,1)}{\partial y}=\int _0^{1} t^{x-1} (-\ln(1-t))\mathop{\text{d} x}$$ And by monotone convergence we have $$\lim_{x\rightarrow 0}\frac{\partial B(x,1)}{\partial y}=\int _0^{1} -\frac{\ln(1-t)}{t}\mathop{\text{d} x}$$ Finally, we have \begin{align} \lim_{x\rightarrow 0}\frac{\partial B(x,1)}{\partial y}&= \lim_{x\to 0}xB(x,y) \lim_{x\to 0}\frac{( \psi(y) - \psi(x+y))}{x}\\ &=1 \cdot(-\psi^{(1)}(1))\\ &=\frac{\pi^2}{6} \end{align} So we conclude.

Answer 5

It is possible to use Feynman’s Trick. The right substitution in this case is inside the natural log

$$I(z)=\int\limits_0^1dx\,\frac {\log(1-zx)}x$$

So when we differentiate, we get

$$I’(z)=-\int\limits_0^1dx\,\frac {1}{1-zx}$$

Can you complete the rest?