Can a Sigma Sum be Integrated with Fubini's Theorem?

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SUMMARY

The discussion centers on integrating a sigma sum using Fubini's Theorem. It is established that integration can be performed term by term for sigma sums, but caution is advised when dealing with infinite sums that do not converge uniformly. A specific example is provided using the function u_m(x) = mxe^{-mx^2} and the series \sum_{n=1}^{\infty} f_n(x), illustrating that \int_0^1 \sum_{n=1}^{\infty} f_n(x) dx does not equal \sum_{n=1}^{\infty} \int_0^1 f_n(x) dx unless conditions for interchangeability are met. The sufficient condition for this interchangeability is a special case of Fubini's Theorem.

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How would you Integrate a Sigma Sum?
 
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If you with a sigma sum means a sum written with the big sigma symbol, you integrate it just like any other sum.
 
arildno said:
If you with a sigma sum means a sum written with the big sigma symbol, you integrate it just like any other sum.
I.e., term by term.
 
thanks.
 
Mark44 said:
I.e., term by term.
Not necessarily, but usually beneficially! :smile:
 
On a related note, an interesting question to consider would be when the orders of a sum and an integral can be interchanged.
 
what?
 
FeDeX_LaTeX said:
On a related note, an interesting question to consider would be when the orders of a sum and an integral can be interchanged.

Exactly. If it's an infinite sum and it doesn't converge uniformly, then we cannot do so term by term. The usual example is from Kresyzig:

Let: u_m(x)=mxe^{-mx^2}

and consider:

\sum_{n=1}^{\infty} f_n(x)

where f_n(x)=u_m(x)-u_{m-1}(x)

then:

\int_0^1 \sum_{n=1}^{\infty}f_n(x) dx\neq \sum_{n=1}^{\infty} \int_0^1 f_n(x)dx
 
Of course you have to make sure the sum and integral are actually interchangeable because this is not always the case. The sufficient condition is a special case of Fubini's theorem.
 

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