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So I have the following definite integral:

${\int_{0}^{1}x(2x^2-1)^{-10}}dx$

I suppose I cannot just integrate over the interval (0,1) because of the discontinuity there. I used the substition method where $t=2x^2-1$ and then the new integral is:

${\int_{-1}^{1}t^{-10}}dt$

which of course still has the discontinuity at $t=0$

So what I did was I broke the integral into two pieces:

${\int_{-1}^{0}t^{-10}}dt$ and ${\int_{0}^{1}t^{-10}}dt$

and used limits to calculate each of them. Of course each of them diverges to $+\infty$ so the whole integral diverges. Is this correct? At first, I made the mistake of not realizing the discontinuity - I calculated the integral without breaking it up into the two integrals mentioned and I got the result $-1/18$. What is the correct answer here? Does the integral exist or not? I know there is something called the Cauchy´s principal value but it´s too advanced for me to understand. Thanks for any help.

kuchejdatomas
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    Your argument seems fine to me. The Cauchy's principal value exists to assign divergent integrals a meaningful value, but they still aren't considered to have a definite value in the normal sense. – Stephen Donovan Oct 31 '21 at 09:09

1 Answers1

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Let $\epsilon<1$, then the function is symmetric on the interval $[-1,-\epsilon]$ and the interval $[\epsilon,1]$. Thus the integral can be rewritten as $\lim_{\epsilon\to0^+}2\int_\epsilon^1t^{-10}dt$, which diverges.

The plot of $f(t)=t^{-10}$ looks as follows:

enter image description here