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How would you deal with such an integral? Any tips will be appreciated!

$$\int_0^{\infty} \left[\prod_{k=1}^K \sum_{j=0}^k a_j x^j\right] f(x) \, dx$$

$a_j$ is a constant not depending on $x$, $f(x)$ is some function of $x$. My question is how to deal with the integral of the product...

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    What is $\mathrm{cst}_j$?2012-10-21
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    Constant strain triangle element?2012-10-21
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    sorry, $\textrm{cst}_j$ is a real number not depending on $x$.2012-10-21
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    You can take the product, the sum and the constant outside the integral.2012-10-21
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    Are you sure you've given this correctly? Every term is going to diverge for nonzero $a_j$.2012-10-21
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    SORRY again! I forgot to write $f(x)$. Let me edit. I am confused!2012-10-21
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    We need some conditions on $f$, don't we?2012-10-21
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    @David Giraudo: Yes, probably. Actually, I am not looking for an exact result. Instead I would like to understand the strategy to attack such an integral2012-10-21
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    @ChrisTaylor *You can take the product .../... outside the integral*... I very much doubt one can.2012-10-21
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    @did: me too :(2012-10-21
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    It seems to me the strategy for analyzing the integral depends on $f$.2012-10-21
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    @did oops, good point.2012-10-21

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Obviously first multiplier is a polynomial of some degree $N=\sum_{k=0}^K k=K(K+1)/2$. Call this polynomial $\sum_{i=0}^{N} b_i x^{i}$. Then you get $$ \int_0^{\infty} \left[\prod_{k=1}^K \sum_{j=0}^k a_j x^j\right] f(x)dx= \int_0^{\infty} \left[\sum\limits_{i=0}^{N} b_i x^{i}\right] f(x) dx= \sum\limits_{i=0}^{N} b_i\int_0^{\infty} x^{i} f(x) dx $$ Hence to attack this integral it is enough to

  • Compute integrals $\int_0^{\infty} x^{i} f(x) dx$ for all $i=0,\ldots,N$
  • Compute coefficients $b_i$, this is very messy.
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    Thanks for your answer. However I have edited the question because I made a mistake. I am confused!2012-10-21
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    Thanks for the edit! Indeed, computed $b_i$ looks very messy!2012-10-21
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    @user7064 even if you get explicit formula for $b_i$ it will be also very messy. So horrible mess is unavoidable in this approach.2012-10-21