Integrating Exponential Functions Multiplied by t^n

In summary, the integral of an exponential function multiplied by any power of t will always converge to a finite value. This can be proven using integration by parts and induction, and it also follows from the fact that the indefinite integral is equal to the factorial of the power of t.
  • #1
Cyrus
3,238
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Its true that if you integrate an exponential function from some time t0 to infinity it will converge to a finite value.

However, is the same true if it is multiplied by say t, t^2, t^3,t^n.


i.e. t*exp(-t) for example.

the exp is decaying to zero faaster than t is, so it goes to zero in the limit. But there are functions that decay to zero but their integral is not finite because the rate of decay is not *fast enough*.

Would the integral of exp multiplied by any power of t ALWAYs converge to a finite number?
 
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  • #2
Yes.

[tex]\int_{t_0}^\infty t^n e^{-t} dt[/tex]

is always finite. You can prove this by induction (hint: use integration by parts).

Here's an interesting factoid for you:

[tex]\int_0^\infty t^n e^{-t} dt = n!.[/tex]
 
  • #3
Yes, using an integration by parts to get an inductive argument. Integrating t*exp(-t) for example will get you

[tex]
-te^{-t} + \int_{t_0}^\infty e^{-t} dt
[/tex]

The left summand will go to 0 and the right integral converges. That establishes that the indefinite integral of t*exp(-t) converges, then since the derivative t^2 is 2t you can establish it again with t^2 and so on (the left summand, t^n*exp(-t) will always go to 0 since t^n = o(e^t) for all n).

Edit: oops, morphism already answered.
 
  • #4
Also consider what happens when you integrate the power series.

t^n*exp(-t) is sufficiently nice that you should be able to interchange the sum and integral.
 

FAQ: Integrating Exponential Functions Multiplied by t^n

1. What is the general formula for integrating exponential functions multiplied by t^n?

The general formula for integrating exponential functions multiplied by t^n is ∫(e^kt)t^n dt = (e^kt)/(k+n) + C, where k is the constant in the exponent and C is the constant of integration.

2. How do you integrate exponential functions multiplied by t^n using substitution?

To integrate exponential functions multiplied by t^n using substitution, first substitute u = e^kt and du = k(e^kt)dt. Next, rewrite the integral in terms of u and du, and then integrate using the power rule. Finally, substitute back in the original variable t to get the final answer.

3. Can you use integration by parts to integrate exponential functions multiplied by t^n?

Yes, integration by parts can be used to integrate exponential functions multiplied by t^n. The general formula for integration by parts is ∫udv = uv - ∫vdu. In this case, let u = t^n and dv = e^kt dt, and then use the power rule and the formula for integrating exponential functions to solve for the integral.

4. What is the difference between integrating exponential functions multiplied by t^n and integrating exponential functions raised to the power of t^n?

The main difference is that when integrating exponential functions multiplied by t^n, the variable t is not raised to a power. This means that integration by parts is typically used, while integrating exponential functions raised to the power of t^n usually involves substitution.

5. Are there any real-world applications for integrating exponential functions multiplied by t^n?

Yes, there are many real-world applications for integrating exponential functions multiplied by t^n. For example, in physics, this type of integration is used to calculate the velocity of an object under constant acceleration. In economics, it can be used to model compound interest. In chemistry, it can be used to model radioactive decay. Overall, this type of integration is useful in any situation where a quantity is increasing or decreasing exponentially over time.

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