Integration of Weird Functions - 1

Calculus Level 5

Define $f(x) = (-1)^{r-1} , \frac{1}{r+1}<x\leq\frac{1}{r}$ for $r=1,2,3...$ and $f(0) = 0$ . Find the value of $\int_{0}^{1} f(x) dx$ . If you think that the function is un-integrable :- enter $1729$ as your answer.

The answer is 0.3863.

1 solution

Arghyadeep Chatterjee
Mar 13, 2020

The set of discontinuity of f(x) in $[0,1]$ is the countable set $S=(0,\frac{1}{2},\frac{1}{3}.......)$ whose set of limit points(derived set) is finite and is infact the singleton set $S'=(0)$ . Hence the function is Riemann Integrable. Now as we can see we can represent $\int_{0}^{1} f(x) dx$ as an infinite summation .

$\int_{0}^{1} f(x) dx = 1(1-\frac{1}{2}) - 1(\frac{1}{2}- \frac{1}{3}) + 1(\frac{1}{3} - \frac{1}{4}) -.......$

$\int_{0}^{1} f(x) dx = (1-\frac{1}{2} + \frac{1}{3} - \frac{1}{4} +.......) + (-\frac{1}{2} + \frac{1}{3} -\frac{1}{4} +.......)$

Using the fact that $\ln(2) = 1-\frac{1}{2}+\frac{1}{3}-\frac{1}{4} +.......$

we have our answer as $2\ln(2) - 1$

It is even easier to observe that the function is Lebesgue integrable over $(0,1)$ , since it is the limit of a sequence of step functions uniformly bounded by the constant function $1$ . Its integral is then $\sum_{r=1}^\infty \frac{(-1)^{r-1}}{r(r+1)} \; = \; 2\ln2 - 1$

Mark Hennings - 1 year, 2 months ago

Yeah. We can view it from Lebesgue integration point of view to show it is integrable. We can also use the concept of set of measure zero to show the function is riemann integrable.

Arghyadeep Chatterjee - 1 year, 2 months ago

The function is Lebesgue integrable because it is a simple function (it takes only two values $\pm 1$ )

Abhishek Sinha - 1 year, 2 months ago

Well, because it is a measurable simple function.

Mark Hennings - 1 year, 2 months ago

More precisely, Borel measurable.

Abhishek Sinha - 1 year, 2 months ago

Integration of Weird Functions - 1

The answer is 0.3863.

1 solution

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