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While evaluating the integral

$$I = \int_0^1 \frac{\ln \left( \displaystyle \sum_{n = 0}^{\infty} x^n \right)}{x} \mathrm{d}x$$

it can be observed that

$$\sum_{n = 0}^{\infty} x^n = \frac{1}{1 - x}$$

for $|x| < 1$. This way, after some steps involving

$$- \ln (1 - x) = \sum_{n = 0}^{\infty} \frac{x^{n + 1}}{n + 1}$$

it turns out that

$$I = \int_0^1 \frac{- \ln(1 - x)}{x} \mathrm{d}x = \frac{\pi^2}{6}$$

However, the equality

$$\sum_{n = 0}^{\infty} x^n = \frac{1}{1 - x}$$

is only valid for $|x| < 1$, while instead the integral includes the point $x = 1$, being it the upper boundary.

According, to this answer, the evaluation of convergence in this special point can not be general, but must be performed according to each specific case.

Then, is it still correct to consider the integrating function as convergent, even if it must also be evaluated in $x = 1$? Why?

BowPark
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    For the integral, the value of the integral at one point is irrelevant (so you can e.g. safely integrate $1/\sqrt{x}$ for $x\in [0,1]$ even though the function is not defined at $0$). – van der Wolf Mar 25 '25 at 11:22
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    This is call an "improper integral." It just means you have to define it via a likit: $$\lim_{s\to 1^-}\int_0^s.$$ Improper is a weird term, because there is nothing wrong with it, it just isn't only an integral, it requires the limit. Thi w is because Riemann integrals have difficulty with it as a sum of Riemann sums. – Thomas Andrews Mar 25 '25 at 11:27

2 Answers2

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As the integrand is unbounded at $x = 1$, we should interpret this as an improper Riemann integral and evaluate as

$$I=\int_0^1 \frac{-\ln(1-x)}{x} \, dx = \lim_{c \to 1-}\int_0^c\frac{-\ln(1-x)}{x} \, dx= \lim_{c \to 1-}\int_0^c\sum_{n=0}^\infty\frac{x^n}{n+1} \,dx$$

To proceed, we invoke two important results. The first is that a power series converges uniformly on a compact subset of its interval of convergence. This justifies the termwise integration

$$I = \lim_{c \to 1-}\sum_{n=0}^\infty\int_0^c\frac{x^n}{n+1}\, dx = \lim_{c \to 1-}\sum_{n=0}^\infty\frac{c^{n+1}}{(n+1)^2}$$

The second result needed is Abels' theorem which justifies interchanging the limit and the sum because the series $\sum_{n=0}^\infty (n+1)^{-2}$ is convergent. Whence,

$$I = \sum_{n=0}^\infty \lim_{c \to 1-}\frac{c^{n+1}}{(n+1)^2}= \sum_{n=0}^\infty \frac{1}{(n+1)^2}$$

RRL
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Firstly, A definite integral evaluated at a point is actually zero with regard to the definition of a definite integral. Hence the inclusion (in your case the point $x=1$) or exclusion of a certain point (be it an end point or an interior point) to or from the interval of integration so defined does not change the value of the integral.

With regard to the integrand function, The interval of integration taken $(0,1)$ or $[0,1)$ or $[0,1]$ or $(0,1]$ will give you the same value in evaluation of the integral.

Taken $0< s < 1$

$$I = \lim_{s\to 1^-}\int_0^s \frac{\ln \left( \displaystyle \sum_{n = 0}^{\infty} x^n \right)}{x} \mathrm{d}x$$

The below identity is valid for all $x \in [0,s]$ and by which $|x| \lt 1$

$$\sum_{n = 0}^{\infty} x^n = \frac{1}{1 - x}$$

Now, I believe you can continue from here. Also, I would like to suggest you taking a glance on Measure theory for further exploration.

vishalnaakar25
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