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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 $\mathrm{cst}_j$? –  userNaN Oct 21 '12 at 10:17
    
Constant strain triangle element? –  FrenzY DT. Oct 21 '12 at 10:18
    
sorry, $\textrm{cst}_j$ is a real number not depending on $x$. –  user7064 Oct 21 '12 at 10:18
    
You can take the product, the sum and the constant outside the integral. –  Chris Taylor Oct 21 '12 at 10:23
    
Are you sure you've given this correctly? Every term is going to diverge for nonzero $a_j$. –  Kevin Carlson Oct 21 '12 at 10:24

1 Answer 1

up vote 0 down vote accepted

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! –  user7064 Oct 21 '12 at 10:27
    
Thanks for the edit! Indeed, computed $b_i$ looks very messy! –  user7064 Oct 21 '12 at 11:17
    
@user7064 even if you get explicit formula for $b_i$ it will be also very messy. So horrible mess is unavoidable in this approach. –  userNaN Oct 21 '12 at 11:19

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