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JG89
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Homework Statement
Prove that [tex] \int_0^{\infty} sin^2(\pi(x + 1/x))dx [/tex] does not exist.
Homework Equations
The Attempt at a Solution
First, we can construct a sequence as follows:[tex] \int_0^{\infty} f(x)dx = \lim_{n \rightarrow \infty} S_n [/tex], where [tex] S_n =\int_0^{1} f(x)dx, \int_0^{2} f(x)dx, \int_0^{3} f(x)dx, \int_0^{4} f(x)dx, ..., \int_0^{n} f(x)dx [/tex]. Now, the sequence S_n converges as n approaches infinity if and only if [tex] |S_m - S_n| < \epsilon [/tex] for all epsilon > 0, provided that there is an integer N such that m,n > N. We set m = n + 1. We then have [tex] |S_{n+1} - S_n| = | \int_0^{n+1} f(x)dx - \int_0^{n} f(x)dx |[/tex].
Now,
[tex]\int_0^{n+1} f(x)dx - \int_0^{n} f(x)dx = \int_n^{n+1} f(x)dx = f(c) [/tex] for some c inbetween n and n + 1 (By the MVT of Integral Calculus). And so [tex] |S_{n+1} - S_n}| = f(c)[/tex]. Now, if |f(c)| < epsilon for sufficiently large c (remember that c goes to infinity as n goes to infinity), then this means that the function approaches a limit of 0, which it obviously doesn't as it's periodic. Thus the integral doesn't converge. QED.How does this proof look?
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