improper integral $$\int_{0}^{2} \frac{1}{(y-1)^{3/2}}\, dy$$
I know it doesn't work when $y = 1$, so I split the integral, right. But then i realized, it doesn't work with $0$ either, as that would leave a $-1$ under the square root.
So how does this work?
On
One guess at what some people might mean by this integral is $$\int_{0}^{2} \frac{1}{(y-1)^{3/2}}\, dy=\lim_{a\to0}\left(\int_0^{1-a} {1\over (y-1)^{3/2}}\,dy+\int_{1+a}^2 {1\over (y-1)^{3/2}}\,dy\right)$$ One often writes things this way in the expectation of a helpful cancellation that leads to a finite limit.
However, the first of these two integrals is purely imaginary and thus tends to $\pm i\infty$, whereas the second is real and tends to $+\infty$. Thus, there is no helpful cancellation and it does not seem possible to give the integral a useful meaning by this route.
$$\int_0^2\frac1{(y-1)^{\frac 3 2}}dy$$
$$=\int_{-1}^1 \frac1{u^{\frac 3 2}}du$$
$$=\int_{-1}^1 u^{-\frac 3 2}du$$
$$= -2u^{-\frac 1 2}|_{-1}^1$$
It looks normal so far, except if you substitute $u=1$ into the expression, you will get imaginary numbers. The reason of that is because the original function is not defined for $y\leq 1$ if your range is the real numbers, as you cannot take square root of negative numbers. (As depicted here http://m.wolframalpha.com/input/?i=%28x-1%29%5E%28-3%2F2%29%3Dy&x=0&y=0 )
So if you insist, you need to break it into two parts:
$$\frac{-2}{\sqrt u} |_{-1}^0 + \frac{-2}{\sqrt u} |_0^1$$
The first part correspond to the imaginary part of the original integral while the second part is the real area. However it does not converge as you see you get $\frac2 0$