# Math Help - Sine Integral Function - Inflection Point

1. ## Sine Integral Function - Inflection Point

Find the coordinates of the first inflection point to the right of the origin:

$Si(x) = \int^x_0 \frac{sin t}{t}dt$

I think that it's obvious that I have to find the second derivative of Si(x) and then set it equal to zero. Here is what I did:

Let $g(x) = Si(x) = \int^x_0 \frac{sin t}{t}dt$

$g'(x) = \frac{dg}{dx} = \frac{d}{dx}[\int^x_0 \frac{sin t}{t}dt = \frac{sin(x)}{x}$ (according to FTC1)

$g''(x) = (\frac{sin(x)}{x})' = (\sin(x)(x^{-1}))' = \cos(x)(x^{-1}) + \sin(x)(-x^{-2}) = \frac{x \cos(x) - \sin(x)}{x^2}$

Assuming all is well to here (big assumption, I know), I get stuck on solving for zero:

$0 = x \cos(x) - \sin(x)$

When I graph this on my calculator, there isn't an obvious solution.

For whatever it's worth, my calculator tells me that the answer to this is approximately $\frac{2}{3}\pi$

Anybody see where I am going wrong or have a suggestion to get me unstuck?

Thanks.

2. I think the only way is to use method of iteration.

EDIT: Or the equation can be rewritten as $x=\tan x$. Draw the curves $y=x$ and $y=\tan x$ on same graph paper; the first point of intersection on the right of the origin will be your answer.

3. Thanks. Do you agree with my approach and my calculation of the 2nd derivative? I think that I can use Newton's method to approximate the root, but I need to make sure that I have the calculation correct up to that point.

4. Yes your calculation is alright.

5. OK, I've worked this problem out and have determined that the x-coordinate of the inflection point is 4.4934 (using Newton's method). Now I need to evaluate Si(4.4934).

I'm applying FTC2 to find this answer:

Let $f(x) = \frac{sin(x)}{x} = F'(x)$

Then, if I can determine F(x), I should be able to find Si(4.4934) = F(4.4394) - F(0). But I don't know how to do this (other than with a calculator).

What do I do? Thanks.

6. It does not have any closed form. You may use the following expansion: ${\rm Si}(x)=\frac{\pi}{2}
- \frac{\cos x}{x}\left(1-\frac{2!}{x^2}+\cdots\right)
- \frac{\sin x}{x}\left(\frac{1}{x}-\frac{3!}{x^3}+\cdots\right)$