Math Problem Statement

Suppose that f is integrable on [0, x] for every x > 0, and that limx→∞ f (x) = a. Prove that lim x→∞ 1 x ∫ x 0 f (t) dt = a . Hint: For N > 0 (to be chosen later) and x > N , write x = N + y and split the integral as ∫ N +y 0 f (t) dt = ∫ N 0 f (t) dx + ∫ N +y N f (t) dt . Take N large and compare the second integral with a · y

Solution

To prove that

limx1x0xf(t)dt=a,\lim_{x \to \infty} \frac{1}{x} \int_0^x f(t) \, dt = a,

given that ff is integrable on [0,x][0, x] for every x>0x > 0 and that limxf(x)=a\lim_{x \to \infty} f(x) = a, we can follow the hint provided and proceed step-by-step.

Step 1: Split the integral

For x>Nx > N, write x=N+yx = N + y, where y>0y > 0, and then split the integral as suggested:

0xf(t)dt=0Nf(t)dt+Nxf(t)dt.\int_0^x f(t) \, dt = \int_0^N f(t) \, dt + \int_N^x f(t) \, dt.

Thus, we have

1x0xf(t)dt=1x(0Nf(t)dt+Nxf(t)dt).\frac{1}{x} \int_0^x f(t) \, dt = \frac{1}{x} \left( \int_0^N f(t) \, dt + \int_N^x f(t) \, dt \right).

Step 2: Deal with the first integral

The first integral 0Nf(t)dt\int_0^N f(t) \, dt is independent of xx and thus becomes negligible as xx \to \infty. Specifically,

limx1x0Nf(t)dt=0.\lim_{x \to \infty} \frac{1}{x} \int_0^N f(t) \, dt = 0.

Step 3: Estimate the second integral

Now, consider the second integral:

Nxf(t)dt=NN+yf(t)dt.\int_N^x f(t) \, dt = \int_N^{N + y} f(t) \, dt.

For large xx, since limtf(t)=a\lim_{t \to \infty} f(t) = a, for sufficiently large NN, f(t)f(t) is close to aa for all t[N,x]t \in [N, x]. More formally, for any ϵ>0\epsilon > 0, there exists some NϵN_\epsilon such that for all t>Nϵt > N_\epsilon, f(t)a<ϵ|f(t) - a| < \epsilon.

Thus, for x>Nϵx > N_\epsilon, we can estimate:

Nxf(t)dtNxadt=Nx(f(t)a)dtNxf(t)adt.\left| \int_N^x f(t) \, dt - \int_N^x a \, dt \right| = \left| \int_N^x (f(t) - a) \, dt \right| \leq \int_N^x |f(t) - a| \, dt.

Since f(t)a<ϵ|f(t) - a| < \epsilon for t>Nϵt > N_\epsilon, we have

Nxf(t)adtϵ(xN)=ϵy.\int_N^x |f(t) - a| \, dt \leq \epsilon (x - N) = \epsilon y.

Therefore,

Nxf(t)dt=a(xN)+O(xN)=ay+O(y).\int_N^x f(t) \, dt = a(x - N) + O(x - N) = a y + O(y).

Step 4: Divide by xx

Now consider the full expression:

1x(0Nf(t)dt+Nxf(t)dt).\frac{1}{x} \left( \int_0^N f(t) \, dt + \int_N^x f(t) \, dt \right).

From the previous steps, we have:

  • 1x0Nf(t)dt0\frac{1}{x} \int_0^N f(t) \, dt \to 0 as xx \to \infty,
  • 1xNxf(t)dt=ay+O(y)x=ayx+O(yx)\frac{1}{x} \int_N^x f(t) \, dt = \frac{a y + O(y)}{x} = \frac{a y}{x} + O\left( \frac{y}{x} \right).

Since y=xNy = x - N and as xx \to \infty, yx1\frac{y}{x} \to 1, we conclude that

1xNxf(t)dta.\frac{1}{x} \int_N^x f(t) \, dt \to a.

Step 5: Conclusion

Therefore,

limx1x0xf(t)dt=0+a=a.\lim_{x \to \infty} \frac{1}{x} \int_0^x f(t) \, dt = 0 + a = a.

Hence, we have shown that:

limx1x0xf(t)dt=a.\lim_{x \to \infty} \frac{1}{x} \int_0^x f(t) \, dt = a.

This completes the proof.


Would you like more details on any part of the proof, or do you have additional questions?

Here are 5 related questions that might help further your understanding:

  1. How do we use the definition of the limit to prove that limxf(x)=a\lim_{x \to \infty} f(x) = a?
  2. What is the significance of the integral test for convergence in this context?
  3. How does the concept of Riemann integrability relate to the problem at hand?
  4. What would happen if f(x)f(x) did not tend to aa as xx \to \infty?
  5. How does the concept of a "local error" in the integral Nx(f(t)a)dt\int_N^x (f(t) - a) \, dt help us bound the difference between the two integrals?

Tip: The key idea in these kinds of proofs is breaking the problem into manageable parts and handling the asymptotic behavior of the function carefully.

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Math Problem Analysis

Mathematical Concepts

Integral Calculus
Limits
Asymptotic Behavior
Riemann Integrability

Formulas

lim x→∞ 1/x ∫ x 0 f(t) dt = a

Theorems

Fundamental Theorem of Calculus
Squeeze Theorem (for limits)

Suitable Grade Level

Grades 12-14 (College/University level)