Indefinite integral approximation technique

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Homework Help Overview

The discussion revolves around the evaluation of the improper integral \(\int_1^\infty \! \frac{(sin(x)+5)}{x^3} \, dx\). The original poster seeks to find two simpler integrals, one that serves as an upper bound and another as a lower bound for the given integral.

Discussion Character

  • Exploratory, Assumption checking

Approaches and Questions Raised

  • Participants explore the idea of simplifying the integral by finding bounds for the sine function. There is discussion on how to establish upper and lower bounds for \(sin(x)\) and how these bounds can be applied to the integral in question.

Discussion Status

Participants are actively engaging with the problem, suggesting that the sine function is bounded between -1 and 1. There is a focus on how these bounds can be utilized to create simpler integrals that are easier to evaluate, although no consensus has been reached on the exact bounds to use.

Contextual Notes

There is an emphasis on finding definite integrals rather than indefinite ones, and the discussion hints at the need for clarity on the bounds of the sine function in relation to the integral.

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Homework Statement


\int_1^\infty \! \frac{(sin(x)+5)}{x^3} \, dx
"Find two simpler integrals, one larger and one smaller."


Homework Equations





The Attempt at a Solution



How could I make this a simpler (ie, solvable) integral? It's been straight forward with other integrals like this:

\int_1^\infty \! \frac{1}{\sqrt{x^4+10}} \, dx \rightarrow \int_1^\infty \! \frac{1}{\sqrt{x^4}} \, dx

Doing this would give an overestimate since the denominator is smaller. But how would I do it for the integral in question?
 
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Can you find upper and lower bounds for sin(x)?
 
Count Iblis said:
Can you find upper and lower bounds for sin(x)?

sin(infinity) and sin(1)? I don't understand what you mean. :redface:
 
Yes, you can. Think of what y=sin(x) is restricted to...
 
Char. Limit said:
Yes, you can. Think of what y=sin(x) is restricted to...

-1 and 1.
 
Exactly. So sin(x) is always less than or equal to 1.

You can use that.
 
And so ? <= sin(x) + 5 <= ?

Filling in the question marks should make two definitie integrals (not indefinite integrals) that are a lot easier to evaluate.
 

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