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How can we convert an unbounded integral to bounded one?

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m.mohamed

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If I have a general integral equation with a semi-bounded limits from zero to infinity and I want to apply this equation on some deterministic values. The integration in this case must be converted to a deterministic integral, but how?

for example, in the equation below, I want to find the integral for a specific range of frequencies.

Unbenannt.jpg
 

c_mitra

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CataM

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Take limit of a --> infinity and substitute the high boundary of the integral with "a"
 

c_mitra

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Because chi(double prime) is not explicit, you need to integrate numerically. Because the integrand has a pole at omega, you need to be careful. If the integral is convergent, fast, then you can look up several tools available free. There can be other problems depending on the pathology of chi(double prime).
 

m.mohamed

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Thanks very much C_mitra. There are no problems with the integration in general but what I meant is how can I apply the same equation on a limited values of omega 'w' for example I need to integrate in the duration [w1,w2] instead of [0, inf.]. Is the integration will be the same in both durations??
 

KlausST

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Hi,

a bounded integral from A to B is calculated with:

integral_unbounded(B) - integral_unbounded(A)

Is this what you needed?

Klaus
 

c_mitra

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... what I meant is how can I apply the same equation on a limited values of omega 'w' for example I need to integrate in the duration [w1,w2] instead of [0, inf.]. Is the integration will be the same in both durations??

I do not understand how you are doing the integration. In general, the integral (from w1 to w2) will be equal to the integral (zero to inf) - integral (zero to w1) - integral (w2 to inf)

OR integral (w1 to w2) = integral ( zero to w2) - integral (zero to w1)

If you change the limits, the sign will change: integral ( w1 to w2) + integral (w2 to w1) = zero.

These are useful if you are using a table or numerical methods
 

CataM

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the integral (from w1 to w2) will be equal to the integral (zero to inf) - integral (zero to w1) - integral (w2 to inf)
c_mitra, Why bother with that?

He already has the integral defined, why not just simply take integral from w1 to w2 ?

What you are saying is useful in case the "integrands" are not defined i.e. he can not perform the integral.
 

m.mohamed

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He already has the integral defined, why not just simply take integral from w1 to w2 ?

OR integral (w1 to w2) = integral ( zero to w2) - integral (zero to w1)

Hi, Thanks c_Mitra & CataM,
what I meant is clear in this eqaution : Untitled.jpg
Is the RHS equal to LHS of the equation ??

by the way, I do not have any infinite values in the measurements, but I can measure only in a specific bound of frequencies [w1,w2].

- - - Updated - - -

take in your account, w1>>0, and w2<< inf.
 

c_mitra

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c_mitra, Why bother with that?...

Sometimes we only a table of values, e.g., error function. In the table we have the integrals listed from 0 to x.

The integrand is not defined at w=+/- w' (it becomes infinite). Only the w=w' will come under the integration limits (other pole is to the left)

- - - Updated - - -

by the way, I do not have any infinite values in the measurements, but I can measure only in a specific bound of frequencies [w1,w2]...

You need to plot f(x) as a function of x. It must becomes reasonably small near zero and near infinity. You can select some range (depending on the accuracy you need) where the integrand is sufficiently small.

Under than condition RHS will be equal to LHS.
 

sudhanwaaa

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First of all,Unbounded integral never becomes bounded.Integral which you are referring as unbounded is actually comes under class of integrals known as Improper integrals.There are several techniques to evaluate them such as DUIS(Differentiation under integral sign),Residue Theorem,Laplace Transform,Fourier Transform(Improper Integrals involving trigonometric entities are also known as Fourier Integrals),Jordan Lemma.For more information you can refer this book :
Calculus with complex numbers :John B. Reade (Chapter 6 and 7)
 

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