1

I am not sure whether to post this question in Signal processing forum or Math forum. Since its related to closed form, I am posting it here. Basically I am trying to compute an expression which by solution, it cannot be computed because no closed form exists. I tried looking over a post here - How can you prove that a function has no closed form integral? which actually says that there are methods to find out whether an expression has a closed form but could not apply to my problem.

Given in the screenshot are the steps for the problem. Highlighted is the expression for which no closed form exists. If you could guide me the way or algorithm to find how to prove that this expression has no closed form, it would be helpful.

enter image description here

sundar
  • 163
  • since $\text{Si}(x)$ has no closed form, it's really left to show that your two $\text{Si}$ terms don't cancel out, $\text{Si}(\cdot) + \text{Si}(\cdot) \ne 0$. For further info, try googling for proofs that $\text{Si}(x)$ has no closed form. – Dando18 Jul 31 '17 at 20:45
  • @Dando18 This makes sense. Because, in the same problem/expression, if we consider this condition, (fc>>(1/T)), the solution exists. So, this ended up in closed form mainly because of the overlapping? If the above is true, is it impossible to compute the addition of two Si functions(unless the limits are given)? – sundar Jul 31 '17 at 20:58
  • 1
    Why do you say $\text{Si}(x)$ has no closed form? It's just as "closed" as any other transcendental function, e.g. $\sin(x)$. – Robert Israel Jul 31 '17 at 22:17
  • @RobertIsrael from one perspective that's true.... however then we approach a slippery slope where we can just give any function conceivable it's own name and conclude all functions have closed forms, which doesn't seem very useful in practice. Somewhere a line has to be drawn (though I generally draw it a little farther our then just the elementary functions) – Brevan Ellefsen Aug 01 '17 at 01:25
  • @Brevan Ellefsen : The key point is the definition of what is a closed form. Usually it is understood as an expression made of a FINITE number of REFERENCED functions. i.e. elementary functions and standard special functions. So, all functions have not closed form. And closed forms are very useful in practice as it is pointed out in a paper for general public : https://fr.scribd.com/doc/14623310/Safari-on-the-country-of-the-Special-Functions-Safari-au-pays-des-fonctions-speciales – JJacquelin Aug 01 '17 at 05:29
  • $\text{Si}$ is a "standard" special function. You'll find it in almost any reference on special functions. – Robert Israel Aug 01 '17 at 20:34

0 Answers0