Examine the convergence of $\int_0^1x^{n-1}\log x dx.$
The solution given is as follows:
$0$ is the only point of infinite discontinuity of the integrand. Let us examine the convergence of $\int_0^{\frac 12}x^{n-1}\log x.$ The integrand is negative in $(0\frac 12]$.Let $f(x)=-x^{n-1}\log x,x\in (0,\frac 12].$ Then $f(x) > 0$for all $x\in (0, \frac 12].$
If $n - 1 > 0,$ the integral $\int_0^{\frac 12}f(x)$ is a proper one, since$\lim_{x\to 0+} x^r\log x= 0,$ for all $r > 0.$
If $n - 1=0,$$0$ is the only point of infinite discontinuity of $f·$Let $m$ be a positive number such that $m + n - 1 > 0.$ Then$\lim_{x\to 0+}x^{m+n-1}\log x=0.$ Therefore $\lim_{x\to 0+} x^m f(x) = 0.$
Let $g(x)=\frac 1{x^m}, x\in (0, \frac1 2].$
Then $g(x) > 0$ for all $x\in (0,\frac1 2].$
Since $lim_{x\to 0+}\frac {f(x)}{g(x)}=0$ and $\int_0^{\frac 12}g(x)dx$ is convergent if $m < 1,$ it follows that $\int_0^{\frac 12}f(x)$ is convergent if $m<1.$
Therefore $\int_0^{1}f(x)$ is convergent if $m<1$ and $m+n-1\gt 0,$ i.e if $1-n\lt m\lt 1,$ i.e if $n>0.$
If $n=0$ the integral reduces to $\int_0^1\frac{\log x}{x}dx.$
$\int_{\epsilon}^1\frac{\log x}{x}dx=-\frac12(\log\epsilon)^2\to -\infty$ as $\epsilon\to 0+$ and therefore $\int_0^1f(x)dx$ is divergent if $n=0.$
If $n < 0,$ then $x^{n-1}\geq x^{-1}$ for all $x\in (0,1].$ Since the integral $\int_0^1\frac{\log x}{x} dx$ is divergent, it follows that, $\int_0^1 f(x)dx$ is divergent.
Hence the given integral is convergent if and only if $n > 0.$
However, there are a couple of thing I don't understand in this solution:
- Problem 1:
If $n - 1 > 0,$ the integral $\int_0^{\frac 12}x^{n-1}\log x$ is a proper one, since$\lim_{x\to 0+} x^r\log x= 0,$ for all $r > 0.$
How is $\int_0^{\frac 12}f(x)$ a proper integral? The function f(x) has 0 as a point of infinite discontinuity, i.e the function is not bounded in $[0,\frac 12].$ Moreover the function $f$ is not even defined at $0.$
- Problem 2:
If $n < 0,$ then $x^{n-1}\geq x^{-1}$ for all $x\in (0,1].$ Since the integral $\int_0^1\frac{\log x}{x} dx$ is divergent, it follows that, $\int_0^1 f(x)dx$ is divergent.
I don't understand why is $\int_0^1 f(x)dx$ divergent just because $\int_0^1\frac{\log x}{x} dx$ is so. If $\int_0^1\frac{\log x}{x} dx$ were convergent then I could have said, $\int_0^1f(x)dx$ is convergent as $\lim_{x\to 0+}\frac{f(x)}{g(x)}=0.$
Any clarifications regarding these two issues will be greatly appreciated.