compagnia04

2022-01-05

Solve the integral:
${\int }_{0}^{2}\frac{dx}{\sqrt{x}\left(x-1\right)}$

encolatgehu

You can break up the integral at any point or points you like. In this case, you could break it up into three integrals: pick a point c strictly between 0 and 1, and consider:
${\int }_{0}^{2}\frac{dx}{\sqrt{x}\left(x-1\right)}={\int }_{0}^{c}\frac{dx}{\sqrt{x}\left(x-1\right)}+{\int }_{c}^{1}\frac{dx}{\sqrt{x}\left(x-1\right)}+{\int }_{1}^{2}\frac{dx}{\sqrt{x}\left(x-1\right)}$
$=\underset{a\to {0}^{+}}{lim}{\int }_{a}^{c}\frac{dx}{\sqrt{x}\left(x-1\right)}+\underset{b\to {1}^{-}}{lim}{\int }_{c}^{b}\frac{dx}{\sqrt{x}\left(x-1\right)}+\underset{d}{\overset{2}{lim}}\frac{dx}{\sqrt{x}\left(x-1\right)}$
The original improper integral exists if and only if each of the three improper integrals exist.

Jenny Bolton

Consider the change of variables $t=\sqrt{x}$. It transforms the interval $\left[0,2\right]$ into $\left[0,\sqrt{2}\right]$ and removes one of the problem zeroes. It is easy to see from the resulting expression that the integral diverges.

karton

Try to solve the integral. You'll find whether it converges. For example, $\int \frac{dx}{\sqrt{x}\left(1-x\right)}$ with $ϵ>0$. Later on, you can study the limit $ϵ\to {0}^{+}$. In this way, you ''kill two birds with a one shot''.