Let's denote the integral from zero to R by I(R).
Then we want to know if
Lim R to infinity of I(R)
exists.
If you have studied the theory of limits and the completeness of the real number system, you know that the limit exists if and only if for every epsilon there exists an X such that for all R1 and R2 larger than X we have:
|I(R1) - I(R2)| < epsilon.
This is the so-called Cauchy criterium.
Now, I(R2) - I(R1) is the integral from R1 to R2. So, what you need to show is that by chosing R1 and R2 sufficiently large, you can make the integral arbitrarily small.
You can do that by substititing
x = sqrt(t) and then by partial integration. You integrate the sin(t) factor while you differentiate the
sin(sqrt(t))/sqrt(t) factor. You can then find an upper bound to the integral (using that the absolute value of sin is always less than or equal to 1), that are strictly decreasing functions of R1 and R2.
Determine whether the following is convergent of divergent.
integral(lower limit=0, upper limit=infinity)of sin(x)sin(x^2)dx
Thanks
1 answer