Can I simply integrate this function?

Question

A function is given by:
$$f(x) = \cases{e^x &$: x < 0$ \\ \cos(x) &: $x \ge 0$ }$$ Find $$\displaystyle{\int_{-\infty}^{\pi/2}f(x)dx}$$

Here comes my question - can I simply break this into two integrals - one from $-\infty$ to $0$ and the second one from $0$ to $\pi/2$ or I firstly have to prove that this function is integrable at $x = 0$, since it is not uniformly defined in any neighborhood of $x = 0$?

Answer 1

Piecewise continuous functions are integrable (assuming the pieces are integrable on the corresponding domains because you're dealing with improper integrals), so you can split the integral up the way you want.

Answer 2

Yes ! If the pieces (sub-functions) of the piecewise function are integrable, then it's integrable as a whole. Specifically for this example, I'd add a limit to the integral for $x<0$, since $0$ is not a part of the first branch :

$$\displaystyle{\int_{-\infty}^{\pi/2}f(x)dx} = \lim_{t \to 0^-}\int_{-\infty}^t e^xdx \space + \space \int_0^{π/2}\cos(x)dx$$

Answer 3

Yes of course. $$\displaystyle{\int_{-\infty}^{ \pi/2 }f(x)dx} =\int_{-\infty}^{0}f(x)dx +\int_{0}^{ \pi/2 }f(x)dx= \int_{- \infty}^{0}e^xdx +\int_{0}^{ \pi/2 } (cosx) dx =2 $$