The senoidal integral function it's defined like $$S_{i}= \int_{0}^x \frac{\sin t}{t}dt.$$ Because, $\frac{\sin t}{t} \rightarrow 1$ when $t \rightarrow 0$, then $Si(x)$ is well defined for all $t$ and is continuous.
a) Draw the graph for $Si$
b) For wich $x$ this function have local maximums?
c) Find the coordinates for the first inflextion points on the right of the origin.
Solution for b). For the test of first and second derivate plus the First Theorem of Calculus we have that $$Si'(t)= \frac{\sin t} {t}$$ and $$Si''(t)= \frac{ t\cos t-\sin t }{t^{2}}.$$ So, $Si'(t) =0 \iff \frac{\sin t} {t} =0 \iff \sin t=0 \iff t=\pi n$, for $n \in \mathbb{Z} \backslash{0}$. Then, $$Si''(\pi n) = \frac{\pi n cos(\pi n) - \sin(\pi n)}{(\pi n)^{2}}=\frac{\cos \pi n}{\pi n}$$ $t$ is a maximum local point for $Si(x)$ if $Si''(t)<0$, and this happen when $\frac{\cos \pi n}{\pi n}<0$. We have two cases
- Case 1. If $n>0$ then $\pi n>0$ and $\cos \pi n<0$, so $t= (2k-1)\pi$ for $k=1,2,...$.
- Case 2. If $n<0$ then $\pi n <0$ and $\cos \pi n >0$, so $t=-2\pi k$ for $k=1,2,...$.
In conclusion, $Si(x)$ have for local maximum points $x= (2k-1)\pi$ and $x=-2\pi k$ for $k=0,1,2,...$.
Am I right?