give me a formal proof of comparison test for improper integrals

Question

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

Answer 1

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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Added in reply to comment below:

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 ∫_a^b f(x) dx ≥ ∫_a^b g(x) dx.

Proof: Divide the interval [a,b] into n subintervals I_k of equal lengthfor k = 1,...,n, and pick an x_k from the k^th interval.

Then the integrals can be approximated by Riemann sums,

Sum[|I_k|*f(x_k)] and Sum[|I_k|*g(x_k)].

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

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 I_k = (x_k-1, x_k).

Let F_k = ∫_Ik f(x) dx and G_k = ∫_Ik g(x) dx.

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

∫_c^d f(x) dx ≥ ∫_c^d g(x) dx.

Taking the limit c → x_k-1 and d → x_k gives that F_k ≥ G_k

Since ∫_a^b f(x) dx = Sum(F_k) and ∫_a^b g(x) dx = Sum(G_k), you get that

∫_a^b f(x) dx ≥ ∫_a^b 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.