1. ## Integral Evaluation

I can't quite figure out how to evaluate this one. Can somebody suggest something?

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

I thought that I could just use FTC2, but then I have to figure out the antiderivative of $(\sin t)(t^{-1})$. I don't know how to find that.

Thanks for any help!

2. The antiderivative is Si(x), which is not an finite elementary antiderivative.

3. Originally Posted by joatmon
I can't quite figure out how to evaluate this one. Can somebody suggest something?

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

I thought that I could just use FTC2, but then I have to figure out the antiderivative of $(\sin t)(t^{-1})$. I don't know how to find that.

Thanks for any help!
This function is well known and does not have an elemetry antiderivative.
Trigonometric integral - Wikipedia, the free encyclopedia

What is the exact wording of your question. If you need to find its derivative the Fundamental theorem of calculus will do the job!

4. This the very last part of a long, multi-part problem. I am trying to find the first positive inflection point to Si(x). I took the second derivative of Si(x), set this equal to zero, and used Newton's method to find the x-coordinate of the inflection point (which was 4.4934). All I have left to do is to find the y coordinate. I can do this easily with a graphing calculator or Mathematica, but I think that we have to do it the long way.

Is there an easier approach?

Thanks.

5. Originally Posted by joatmon
This the very last part of a long, multi-part problem. I am trying to find the first positive inflection point to Si(x). I took the second derivative of Si(x), set this equal to zero, and used Newton's method to find the x-coordinate of the inflection point (which was 4.4934). All I have left to do is to find the y coordinate. I can do this easily with a graphing calculator or Mathematica, but I think that we have to do it the long way.

Is there an easier approach?

Thanks.

Since it does not have an elemetry antiderivative you a few options.

First you could use a quadrature rule to estimate the value of the integral

Simpsons rule, Trapezoid, midpoint or evena Riemann sum.

2nd an the more commin if you know infinite series

$\displaystyle \frac{\sin(t)}{t}=t^{-1}\sum_{n=0}^{\infty}\frac{(-1)^nt^{2n+1}}{(2n+1)!}=\sum_{n=0}^{\infty}\frac{(-1)^nt^{2n}}{(2n+1)!}$
You can integrate this term by term to get an infinite series representation of the function and use the Alternating Series test to bounded the error as small as you wish.