Evaluate INT (5+cosx)e^-x from 0 to infinity

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Homework Help Overview

The discussion revolves around evaluating the improper integral of the function (5 + cos(x))e^(-x) from 0 to infinity. Participants are exploring convergence and methods for evaluating the integral, which involves concepts from calculus and integration techniques.

Discussion Character

  • Exploratory, Mathematical reasoning, Assumption checking

Approaches and Questions Raised

  • Participants discuss splitting the integral into two parts and comparing it with simpler functions to establish convergence. There are mentions of using integration by parts and inequalities involving the cosine function to support their reasoning.

Discussion Status

Several participants have offered different methods for approaching the problem, including comparisons with known convergent integrals. There is an ongoing exploration of the best approach to demonstrate convergence and evaluate the integral, with no explicit consensus reached.

Contextual Notes

Some participants note the non-negativity of the integrand and the implications of this property for convergence. The discussion also reflects on the constraints of the original problem and the assumptions made regarding the behavior of the cosine function.

IntegrateMe
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ok so i got (5+cos x)/ex and i compared that with 1/ex

(they're both going from 0 to infinity).

Turns out that the integral of 1/ex from 0 to infinity converges to 1. But i don't know how to prove that our original function converges as well (which is the answer). Anyone care to help?
 
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Split the integral into: [tex]5\int_0^\infty{e^{-x}}dx+\int_0^\infty{cosxe^{-x}}dx[/tex]

and since you know the first one is equal to 5, just use integration by parts to find the second and then add 5 to it.
 
You could also compare your original integral with 6/e^x. Which is larger?
 
If you're just trying to prove that the integral converges, the fact that [itex]\int_0^{\infty}\exp{(-x)}\mathrm{d}x[/itex] converges is enough to prove that the second integral in Mentallic's post converges. For example, [itex]|\cos{(x)}| \leq 1[/itex] from which it follows that [itex]\exp{(-x)}|\cos{(x)}| = |\exp{(-x)}\cos{(x)}| \leq \exp{(-x)}[/itex] and then using this inequality we can reduce the convergence of the second integral to mere trivialities. However, if you actually need to evaluate the integral, Mentallic's method is definitely the best way to go.

Edit: I would follow Dick's advice since his suggestion is considerably simpler than mine.
 
jgens said:
If you're just trying to prove that the integral converges, the fact that [itex]\int_0^{\infty}\exp{(-x)}\mathrm{d}x[/itex] converges is enough to prove that the second integral in Mentallic's post converges. For example, [itex]|\cos{(x)}| \leq 1[/itex] from which it follows that [itex]\exp{(-x)}|\cos{(x)}| = |\exp{(-x)}\cos{(x)}| \leq \exp{(-x)}[/itex] and then using this inequality we can reduce the convergence of the second integral to mere trivialities. However, if you actually need to evaluate the integral, Mentallic's method is definitely the best way to go.

Edit: I would follow Dick's advice since his suggestion is considerably simpler than mine.

Nah, it's not that much simpler. Mine only works because (5+cos(x)) is nonnegative. If it were (1+cos(x)) IntegrateMe would definitely want to use your comparison.
 
First break the integrand into twp parts as shown by Mentallic.You can reduce the pain of integration by writing cos(x) as Re[exp(ix)];next solve the integration,which is now trivial.Then,take the real part of the resulting indefinite integral and put the limits.The lower limit gives you finite value and the upper limit gives you zero for the e^-x factor.
 

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