When reviewing improper integrals, I came across the following explanation;
To find the integral
$$\int_{2}^{5}\frac{1}{\sqrt{x-2}} \,dx,$$
we have to proceed as
$$\int_{2}^{5}\frac{1}{\sqrt{x-2}} \,dx=\lim_{t\rightarrow2^{+}}\int_{t}^{5}\frac{1}{\sqrt{x-2}} \,dx=\lim_{t\rightarrow2^{+}}2\sqrt{x-2}\,\Big|_t^5
=2\sqrt{3}$$
since the function $f(x)=1/\sqrt{x-2}$ has a singularity at $x=2$.
But in this case, an antiderivative of $f$, $F(x)=2\sqrt{x-2}\,$, is continuous on $[2, 5]$, so can't we just use the FTC and say as follows?
$$\int_{2}^{5}\frac{1}{\sqrt{x-2}} \,dx=2\sqrt{x-2}\,\Big|_2^5=2\sqrt{3}$$
I am studying for the GRE subject test, and I saw a couple of other examples that tell you to use the limit for the limit of integration.
Another example is
$$\int_{0}^{\infty}\frac{e^{-\sqrt{x}}}{\sqrt{x}} \,dx.$$
Here, the explanation is that since $g(x)=e^{-\sqrt{x}}/\sqrt{x}$ has a discontinuity at $x=0$, we have to treat the lower limit as $$\lim_{a\rightarrow0^{+}}\int_{a}^{\infty}\frac{e^{-\sqrt{x}}}{\sqrt{x}} \,dx.$$
But an antiderivatie of $g$ is continuous at $x=0$, so I don't know why this is necessary.
Best Answer
The FTC guarantees the existence of an antiderivative when the integrand is everywhere continuous. If the integrand has a singularity anywhere on the domain $[a,b]$ then it may or may not have an antiderivative continuously definable on $[a,b]$. If such antiderivative does exist, say $F$ of $f$, then (e.g.) $\lim_{c\to0^+}\int_c^1f=\lim_{c\to0^+}F(1)-F(c)=F(1)-F(0)$ since $F$ is continuous. All is well: using antiderivative is a valid way to compute these integrals for continuous integrands.
The trouble lies in how an integral is defined. In Riemann integration, $\int_a^bf$ only makes sense if $f$ is every defined and finite on $[a,b]$ (and integrable…). So, should there be any singularity, we make a new definition by taking limits around the singularities. Though antiderivative are useful for computation, we can’t use antiderivatives in the definition since they may not necessarily exist: we may want to integrate a non-continuous function, e.g.
Consider $f(x)=1/x$. On $(0,1]$ it has an antiderivative $\ln x$. But no value of $\ln(0)$ will make $\ln:[0,1]\to\Bbb R$ continuous! And, as one might expect from this, $\int_0^1f$ does not converge as an improper integral.