# Can some improper integrals be directly evaluated?

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?

• Equation $(2)$ is just shorthand for $(1)$. – Infiaria Mar 8 '19 at 14:10
• I apologize for the lack of clarification. I mean taking the antiderivative of $f(r)$ and then evaluating the limits, i.e. $F(r)-F(0)$. Let me edit the question. – Joe Mar 8 '19 at 14:28

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 $$\delta>0$$).
• Is there any difference between $r'$ and $r$ – Joe Mar 8 '19 at 14:55
• @Joe You can't use the same variable in an integration limit and as the variable you integrate over. For example, $x^2=\int_0^x 2tdt$ is true but "$\int_0^x2xdx$" is nonsense. – J.G. Mar 8 '19 at 14:56
• Why is it so? Both will give $x^2$. Is it a convention to not get confused, or is it strictly prohibited like division by zero? – Joe Mar 8 '19 at 15:00
• @Joe It's just a meaningless expression. I can write $\int_0^y2xdx$, but I can't write $\int_0^x2xdx$ because the expression simultaneously says "$x$ is a dummy integration variable" and "$x$ is one of the limits of that variable's range". – J.G. Mar 8 '19 at 15:11
• @Joe $\int_\delta^r f(r^\prime) dr^\prime$ is shorthand for $\int_{r^\prime\in[\delta,\,r]} f(r^\prime) dr^\prime$, whereas we can't specify a range with the condition $r\in[\delta,\,r]$. – J.G. Mar 8 '19 at 15:27

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 $$0 but is unbounded on $$[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 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 $$0 and unbounded on $$[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}$$