Math Problem Statement

An integral formula for the factorials. For n = 0, 1, . . ., consider In := lim T →∞ ∫ T 0 tne−t dt . Prove by induction that these limits exist, and find their values

Solution

To prove that the limits In=limT0TtnetdtI_n = \lim_{T \to \infty} \int_0^T t^n e^{-t} \, dt exist and to find their values, we can proceed step by step.

Step 1: Base Case

Start with n=0n = 0. The integral becomes: I0=limT0TetdtI_0 = \lim_{T \to \infty} \int_0^T e^{-t} \, dt This is a straightforward integral: 0Tetdt=[et]0T=1eT.\int_0^T e^{-t} \, dt = \left[-e^{-t}\right]_0^T = 1 - e^{-T}. As TT \to \infty, eT0e^{-T} \to 0, so: I0=1.I_0 = 1. Thus, the base case holds: I0=1I_0 = 1.

Step 2: Inductive Step

Now, assume that for some n=kn = k, the following is true: Ik=limT0Ttketdtexists and is finite.I_k = \lim_{T \to \infty} \int_0^T t^k e^{-t} \, dt \quad \text{exists and is finite}. We want to show that the integral for n=k+1n = k+1 also exists and find its value.

Integral for n=k+1n = k+1

Consider the integral: Ik+1=limT0Ttk+1etdt.I_{k+1} = \lim_{T \to \infty} \int_0^T t^{k+1} e^{-t} \, dt. We can integrate by parts, where we let:

  • u=tk+1u = t^{k+1}, so that du=(k+1)tkdtdu = (k+1)t^k \, dt,
  • dv=etdtdv = e^{-t} \, dt, so that v=etv = -e^{-t}.

Using the integration by parts formula: udv=uvvdu,\int u \, dv = uv - \int v \, du, we have: 0Ttk+1etdt=[tk+1et]0T+(k+1)0Ttketdt.\int_0^T t^{k+1} e^{-t} \, dt = \left[ -t^{k+1} e^{-t} \right]_0^T + (k+1) \int_0^T t^k e^{-t} \, dt. Now, evaluate the boundary terms: [tk+1et]0T=Tk+1eT+0=Tk+1eT.\left[ -t^{k+1} e^{-t} \right]_0^T = -T^{k+1} e^{-T} + 0 = -T^{k+1} e^{-T}. As TT \to \infty, Tk+1eT0T^{k+1} e^{-T} \to 0 because the exponential decay of eTe^{-T} dominates any polynomial growth of Tk+1T^{k+1}. So, the boundary term vanishes: limTTk+1eT=0.\lim_{T \to \infty} -T^{k+1} e^{-T} = 0. Thus, we are left with: Ik+1=(k+1)limT0Ttketdt=(k+1)Ik.I_{k+1} = (k+1) \lim_{T \to \infty} \int_0^T t^k e^{-t} \, dt = (k+1) I_k.

Step 3: Conclusion

By induction, we have shown that In=n!I_n = n! for all nn. Specifically:

  • I0=1=0!I_0 = 1 = 0!,
  • I1=1=1!I_1 = 1 = 1!,
  • I2=2=2!I_2 = 2 = 2!,
  • and so on.

Therefore, we conclude: In=limT0Ttnetdt=n!for alln=0,1,2,I_n = \lim_{T \to \infty} \int_0^T t^n e^{-t} \, dt = n! \quad \text{for all} \quad n = 0, 1, 2, \dots


Would you like a more detailed explanation of any of the steps or further examples? Here are 5 related questions you might find interesting:

  1. How does the method of integration by parts generalize for higher-order integrals?
  2. How does the exponential decay ete^{-t} affect the convergence of integrals as TT \to \infty?
  3. What is the relationship between the Gamma function and factorials in terms of integrals?
  4. Can we derive the factorial expression using a recurrence relation instead of induction?
  5. How does the integral 0tnetdt\int_0^\infty t^n e^{-t} \, dt relate to the Gamma function?

Tip: Whenever you're dealing with limits of integrals involving exponential decay, check if integration by parts can simplify the process and eliminate troublesome terms.

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Math Problem Analysis

Mathematical Concepts

Integral Calculus
Induction
Factorials
Exponential Functions

Formulas

I_n = lim(T→∞) ∫_0^T t^n e^−t dt
Integration by Parts formula: ∫ u dv = uv - ∫ v du

Theorems

Inductive Proof
Factorial Formula

Suitable Grade Level

Grades 11-12 / Undergraduate