Integration Work Check: Solving for e^(-nx) x^(s-oo-2) dx

In summary, the conversation discusses an integral with symbols and limits that cannot be read by a webreader. The question asks if the value of the integral can be confined to a smaller interval by binding a constant value to the equation. The goal is to determine if the integral converges and, if so, how small the interval of the solution can be.
  • #1
rman144
35
0
Can someone confirm that the following is correct:

S [e^(-nx) ][x^(s-oo-2)] dx , limits from x=1 to x=infinity, n is some real integer>=1, and 0<Re(s)<1

What I want to know is, because the equation vanishes for all "x" in the interval x=1 to x=infinity, excluding x=1, can I just bind (e^(-nx)) to e^(-n(1)), leaving:

S [e^(-nx) ][x^(s-oo-2)] dx=(e^(-n))*S [x^(s-oo-2)] dx

Any help would be much appreciated.
 
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  • #2
1. There are symbols in that my webreader cannot read.
2. I don't know what "s-∞-2" could possibly mean.
3. I don't know what "bind (e^(-nx)) to e^(-n(1))" means.
 
  • #3
Sorry about the confusion:

First, the s-oo-2 is read as "s" (the variable s) minus infinity minus 2.

Next, when I say bind it to a constant, I mean:

Say you have:

S f(x)*g(x) dx, limits from x=x1 to x=x2

Now if you know that the equation f(x) on the interval x1 to x2 is at most f(k1) and at least f(k2), then the integral:

S f(x)*g(x) dx

Will have a maximum absolute value of:

abs( S f(k1)*g(x) dx )

And a minimum of:

abs( S f(k2)*g(x) dx)

Now what I'm asking is, because within my original equation, the only value in the interval that doesn't go to zero is when x=1, as a result, can I repeat the above process for e^(-nx) and "bind" the value to the interval:

e^-n<=e^(-nx)<=e^-n

Which implies:

e^(-nx)=e^(-n)


I'm not trying to solve the integral; I'm merely attempting to establish whether or not it converges, and if so, how small an interval can I confine the possible solution of the actual integral to.
 
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  • #4
rman144 said:
Sorry about the confusion:

First, the s-oo-2 is read as "s" (the variable s) minus infinity minus 2.
Yes, even I could read[/i] it that way but what does it MEAN? Since "infinity" is not a number, that makes no sense to me. Is it intended as a limit? And if so what could "infinity minus 2" mean?

Next, when I say bind it to a constant, I mean:

Say you have:

S f(x)*g(x) dx, limits from x=x1 to x=x2

Now if you know that the equation f(x) on the interval x1 to x2 is at most f(k1) and at least f(k2), then the integral:

S f(x)*g(x) dx

Will have a maximum absolute value of:

abs( S f(k1)*g(x) dx )

And a minimum of:

abs( S f(k2)*g(x) dx)

Now what I'm asking is, because within my original equation, the only value in the interval that doesn't go to zero is when x=1, as a result, can I repeat the above process for e^(-nx) and "bind" the value to the interval:

e^-n<=e^(-nx)<=e^-n

Which implies:

e^(-nx)=e^(-n)


I'm not trying to solve the integral; I'm merely attempting to establish whether or not it converges, and if so, how small an interval can I confine the possible solution of the actual integral to.
 

FAQ: Integration Work Check: Solving for e^(-nx) x^(s-oo-2) dx

What is integration work check?

Integration work check is a method used in calculus to solve for the value of a definite integral. It involves manipulating the integrand (the function being integrated) to make it easier to integrate, and then using the fundamental theorem of calculus to find the solution.

How do I solve for e^(-nx) x^(s-oo-2) dx using integration work check?

To solve for this integral using integration work check, you would first use algebraic manipulation to rewrite the integrand in a form that is easier to integrate. This may involve factoring, substitution, or using trigonometric identities. Once the integrand is simplified, you can then use the fundamental theorem of calculus to find the solution.

What is the fundamental theorem of calculus?

The fundamental theorem of calculus states that if a function f(x) is continuous on an interval [a, b], and F(x) is the antiderivative of f(x), then the definite integral of f(x) from a to b is equal to F(b) - F(a). In other words, the value of a definite integral can be found by evaluating the corresponding antiderivative at the limits of integration.

Why is integration work check important in calculus?

Integration work check is important in calculus because it allows us to find the value of definite integrals, which are used in many applications in mathematics and science. These applications include finding areas under curves, calculating volumes of three-dimensional shapes, and solving differential equations.

What are some tips for using integration work check effectively?

Some tips for using integration work check effectively include practicing algebraic manipulation and being familiar with common integration techniques, such as substitution and integration by parts. It is also helpful to carefully check your work and use proper notation to avoid mistakes. Additionally, understanding the properties of integrals, such as linearity and the power rule, can make the process easier.

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