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# give me a formal proof of comparison test for improper integrals

please i need a formal proof of comparison test for improper integrals..

Hint: Use the fact that an improper integral is a limit of a proper integral as one or both bounds approach either ±∞ or values where the function is discontinuous. Or could be that a value inside an interval where the function is discontinuous.

If you also want to start by proving of the comparison test for proper integrals, you can do so by interpreting a definite integral as a limit of a Riemann sum.

I hope this helps.

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Here is an outline. You will need to fill in the details since that will give you an opportunity to practice writing formal proofs.

First you prove a lemma for proper integrals since an improper integral is a limit of a proper integral.

Lemma:

If f(x) and g(x) are continuous in a finite interval [a,b]

with f(x) ≥ g(x) ≥ 0 then ∫ab f(x) dx ≥ ∫ab g(x) dx.

Proof: Divide the interval [a,b] into n subintervals Ik of equal lengthfor k = 1,...,n, and pick an xk from the kth interval.

Then the integrals can be approximated by Riemann sums,

Sum[|Ik|*f(xk)] and Sum[|Ik|*g(xk)].

Term by term comparison shows that Sum[|Ik|*f(xk)] ≥ Sum[|Ik|*g(xk)].

Letting n → ∞, the sums approach the integrals and the lemma follows

Proof of the comparison test:

Let f(x) ≥ g(x) ≥ 0 over (a,b) wherever both f and g are continuous. Here a could be -∞, and b could be +∞.

Also assume there may be x values where f or g is discontinuous and let them partition (a,b) into intervals Ik = (xk-1, xk).

Let Fk = ∫Ik f(x) dx and Gk = ∫Ik g(x) dx.

Then for any subinterval [c,d] of Ik, our lemma guarantees

cd f(x) dx ≥ ∫cd g(x) dx.

Taking the limit c → xk-1 and d → xk gives that Fk ≥ Gk

Since ∫ab f(x) dx = Sum(Fk) and ∫ab g(x) dx = Sum(Gk), you get that

ab f(x) dx ≥ ∫ab g(x) dx.

Finally, using this inequality, if the integral in f converges, so does the one in g, while if the one in g diverges, so does the one in f.