Exponential function

  • Thread starter Cyrus
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  • #1
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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?
 

Answers and Replies

  • #2
morphism
Science Advisor
Homework Helper
2,015
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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
1,425
1
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
655
3
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.
 

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