Can some improper integrals be directly evaluated?

Question

Consider the following improper integral:

$\displaystyle \lim\limits_{\delta{r} \to 0} \int^r_{\delta r} f(r) dr \tag{1}$

where $f(r)$ is finite everywhere and $f(0)=$ not defined.

Then, will taking the antiderivative of $f(r)$ and then evaluating the limits

$\displaystyle \left[ \int f(r) dr \right]_r-\left[ \int f(r) dr \right]_0 \tag{2}$

yield the same result always? Why?

Answer 1

I'll write $\delta$ instead of $\delta r$, since we're not denoting a small change in $r$, but a small value of it. Then an antiderivative $F$ of $f$ satisfies $$\lim_{\delta\to0}\int_\delta^r f(r^\prime)dr^\prime=\lim_{\delta\to0}(F(r)-F(\delta))=F(r)-\lim_{\delta\to0}F(\delta),$$so the existence condition is that $F(r),\,\lim_{\delta\to0}F(\delta)$ exist and their difference isn't an indeterminate form such as $\infty-\infty$. Bear in mind in particular that:

• $\lim_{\delta\to0}F(\delta)$ may be rewritten as $F(0)$ iff $F$ is continuous at $0$;
• These are constraints on $F$, not $f$ (consider for example $f(r^\prime):=r^{\prime-1/2}$ with some $\delta>0$).

Answer 2

Your post needs some clarification. When you write "$f(r) $ is finite everywhere and $f(0)$ is not defined" what you really mean is that $f$ is bounded because if $f $ is defined at $r$ then $f(r) $ is finite.

In case when $f$ is bounded, the integral $\int_{0}^{r}f(t)\,dt$ is proper. The need for an improper integral arises only when either the integrand is unbounded or the interval of integration is unbounded.

Assuming then that $f$ is bounded and Riemann integrable on $[0,r]$ and possesses an anti-derivative $F$ on $(0,r)$ then both $\lim_{x\to r^{-}} F(x) $ and $\lim_{x\to 0^{+}}F(x)$ exist and $$\int_{0}^{r}f(t)\,dt=\lim_{x\to r^{-}} F(x) - \lim_{x\to 0^{+}}F(x)$$ A non-trivial example is $$f(x) = 2x\sin(1/x)-\cos(1/x)$$ with anti-derivative $$F(x) =x^2\sin(1/x)$$ and we have $$\int_{0}^{1}(2x\sin(1/x)-\cos(1/x))\,dx=\sin 1-\lim_{x\to 0^{+}}F(x)=\sin 1$$

In case when $f$ is bounded on every interval of type $[h, r] $ for $0<h <r$ but is unbounded on $[0,r]$ then the integral $\int_{0}^{r}f(t)\,dt$ is improper and defined by $$\int_{0}^{r}f(t)\,dt=\lim_{h\to 0^{+}}\int_{h}^{r}f(t)\,dt$$ Assume that $f$ possesses an anti-derivative $F$ on $(0,r)$ and $f$ is Riemann integrable on every interval $[h, r] $ with $0<h<r$ then $\lim_{x\to r^{-}} F(x) $ exists and we have $$\int_{h} ^{r} f(t) \, dt=\lim_{x\to r^{-}} F(x) - F(h) $$ and the improper integral $\int_{0}^{r}f(t)\,dt$ exists if and only if $\lim_{h\to 0^{+}}F(h)$ exists and then we have $$\int_{0}^{r}f(t)\,dt=\lim_{x\to r^{-}} F(x) - \lim_{h\to 0^{+}}F(h)$$ Consider an example $$f(x) =\dfrac{1}{\sqrt{2x-x^2}}$$ which is bounded and Riemann integrable on every interval of type $[h, 1]$ with $0<h<1$ and unbounded on $[0,1]$. It also possesses an anti-derivative $F(x) =-\arcsin(1-x)$ on $(0,1)$. Under these circumstances the limit $\lim_{x\to 1^{-}}F(x)$ must exist and clearly the limit is $0$. By our luck the limit $\lim_{x\to 0^{+}}F(x)$ also exists and equals $-\pi/2$ and thus the improper integral $\int_{0}^{1}f(t)\,dt$ exists and has value $\pi/2$ ie $$\int_{0}^{1}\frac{dx}{\sqrt{2x-x^2}}=\frac{\pi}{2}$$