Maximum and minimum value of $\theta(x)=\int_0^x\frac{\sin z}{z}dz$

Question

Let $\theta(x)=\int_0^x\frac{\sin z}{z}dz$, x>0. Then $\theta(x)$ has

(A) maximum for $x = nπ$, n = 2, 4, 6, . . . . . . .

(B) minimum for $x = nπ$, n = 1, 3, 5, . . . . . . .

(C) maximum for $x = nπ$, n = 1, 3, 5, . . . . . . .

(D) minimum for $x = nπ$, n = 2, 4, 6, . . . . . . .

My approach is as follow $\theta(x)=\int_0^x\frac{\sin z}{z}dz$

$\theta'(x)=\frac{\sin x}{x}$

$\theta''(x)=\frac{x\cos x-\sin x}{x^2}$

I cannot proceed from here

Answer 1

$\theta'(x)=0$ iff $x =n\pi$ for some positive integer $n$. When $x =n\pi$ we have $\theta''(x)=\frac {n\pi (-1)^{n}} {n^{2}x^{2}}$ and this is positive for $n$ even , negative for $n$ odd. Hence $\theta$ has local maximum at the points $n\pi$ with $n$ odd and it has local minimum at the points $n\pi$ with $n$ even. So A) and B) are false, C) and D) are true.

Answer 2

I'm going to give you a more graphical approach, though it is the same if done analytically.

Imagine the function $f(x) = \frac{\sin x}{x}$. The function is basically a wave with amplitude $\dfrac{1}{x}$. It will look like a \sin wave, and have the same zeros, but the amplitude will keep damping. At points where $\sin x = +1,-1$, the wave will touch the graphs of $y=\dfrac{1}{x}, \dfrac{-1}{x}$.

Now when you integrate this function, you find the area under the curve from $0$ till a particular point. Then the options will make sense, and the answer will be quite easy to ensure.