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I think there is some mistake in the steps shown below. My objection is while integrating the function $f(x)$ we cannot directly put $1$ although the series when integrated has been substituted by $1$ directly. Kindly help me understand.

$\displaystyle\int_0^1f(x)dx=\displaystyle\int_0^1\frac{x}{1-x}dx=\displaystyle\int_0^1x+x^2+x^3+\ldots\infty dx$

$=\left.\frac{x^2}{2}+\frac{x^3}{3}+\frac{x^4}{4}+\ldots\infty\right\vert_0^1$

$=\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+\ldots\infty$

Aiden Chow
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Akash Gautama
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1 Answers1

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It is very easy to make this rigorous: $\int_0^{1} f(x) dx\geq \int_0^{1} [x+x^{2}+...+x^{N}]dx=\frac 1 2+\frac 1 3+...+\frac 1{N+1}$ for each $N$ and $ \frac 1 2+\frac 1 3+...+\frac 1{N+1}\to \infty$ so $\int_0^{1} f(x) dx =\infty$.

Kavi Rama Murthy
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  • I understand that the integral is divergent but can we say integral of x/(1-x) from 0 to 1 is 1/2+ 1/3 +1/4 +..... – Akash Gautama Feb 14 '20 at 05:39
  • There is theorem called the Monotone Convergence Theorem which tells us that the steps in the argument are justified. But if you want to avoid the theorem you can use the argument I gave. @AkashGautama – Kavi Rama Murthy Feb 14 '20 at 05:43
  • My point is both integral 0 to 1 of x/(1-x) and 1/2+1/3+1/4+.... are divergent. Can these be equated as suggested in the question. – Akash Gautama Feb 14 '20 at 07:34
  • Yes. Measure theory helps you to easily handle this situation. If $f_i \geq 0$ and measurable for each $i$ then $\int \sum_i f_i d\mu=\sum_i \int f d\mu$ is true without any assumption on finiteness of the two sides. @AkashGautama – Kavi Rama Murthy Feb 14 '20 at 07:38