# Integrability of a piecewise function [closed]

I got this question:

Is the function $g$ defined below is integrable on $[0,1]$?

$$g(x)= \begin{cases} \dfrac{1}{x}\sin\dfrac{1}{x} & \text{if x\neq 0} \\[8pt] 3 & \text{if x=0} \end{cases}$$

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## closed as off-topic by mau, Davide Giraudo, Grigory M, Tom Oldfield, userNaNJun 5 at 21:41

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What kind of integral do you seek? –  ellya Jun 5 at 17:33

## 2 Answers

The easiest way to get around this is to use the Riemann-Lebesgue's theorem: It states that a function $f$ defined on a compact set is Riemann integrable iff it is bounded and the set of points of discontinuities has measure zero. Observe that $[0,1]$ is compact, and the function is unbounded because $f\left(\dfrac{1}{\dfrac{\pi}{2} + 2n\pi}\right) = 2n\pi + \dfrac{\pi}{2} \to +\infty$. So the function is not Riemann integrable.

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It may depend a bit on what kind integral you are using. The Lebesgue integral of $g$ does not exist, while the Riemann integral does. Try substiting $y=1/x$ and you'll see that your integral is the same as

$$\int_1^\infty \frac{\sin(y)}{y} dy$$

The value 3 at 0 is kind of a red herring since the value at a single point does not influence the value of the integral. The integral of $\frac{\sin(y)}y$ is also known as sine integral. There is actually no explicit way to write down an indefinite integral. Note that it looks a lot like the series

$$\sum_{k=1}^\infty \frac{(-1)^k}{k}$$

slowly diminishing in absolute value but changing sign all the time. This series converges, but there is no absolute convergence. Roughly speaking each term in this series corresponds to a "bump" of $\frac{\sin(y)}{y}$.

Now if you want the Lebesgue-integral, then $\left|\frac{\sin(y)}{y}\right|$ needs to be integrable, which it isn't, you can show that by comparing the "bumps" to the divergent series $\sum_{k=1}^\infty \frac{1}{k}$. For the Riemann integral note, that you can calculate the value

$$\int_0^\infty \frac{\sin(x)}{x} dx = \frac{\pi}{2}$$

via the Laplace transform and note that $\int_0^1 \frac{\sin(x)}{x} dx$ is finite, since the function is continuous and bounded.

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Thanks for the answer. –  Saita Jun 6 at 8:44