Skip to main content
R
rdelascasas12-Amethyst
12-Amethyst
May 17, 2013
Solved

Improper integral does not work

  • May 17, 2013
  • 2 replies
  • 4919 views

Dear sirs,

I am trying to obtain the improper integral that appear in the attached file. I cannot find the mistakes that I am having here.

As you can see the variables and constants inside the integral are column matrices.

Any advice will be highly appreciated.

thank you in advance,

Rogelio

Best answer by Werner_E

If you still need the vector argument - both sheets, your original one and the modified one of Alan work, if you properly vectorize and use symbolic evaluation. However calculation time is rather long,

Find attached both sheets appropriatly modified - I just limited the vectors to three elements instead of the original 21 to "speed" up execution time. I haven't tried but I would expect the sheets to evaluate with larger vectors, too (unless it will run out of memory - not sure if this could be an issue).

Remark: The original sheet will take about 35% longer to evaluate compared to Alan's modification!

2 replies

W
Werner_E25-Diamond I
25-Diamond I
May 17, 2013

I would first try with scalar values to make that indefinite integral evaluate.

Then, when applying vectors, you should vectorize your expressions.

A
AlanStevens19-Tanzanite
19-Tanzanite
May 18, 2013
Rogelio de las Casas wrote:
...
I am trying to obtain the improper integral that appear in the attached file.

It's a proper integral - it has limits.

Note Werner's comments.

Also, the upper limit of your integral is too high - it involves numbers greater than Mathcad can handle. However, a lower limit works - you can test convergence by trying a few different upper limits. See attached.

There are improvements you could make to the worksheet. For example you've defined three different alpha functions, but you really only need one to which you pass the appropriate parameters. However, I haven't taken the time to do this in the attached.

Alan

W
Werner_E25-Diamond I
25-Diamond I
May 18, 2013

It's a proper integral - it has limits.

Note Werner's comments.

But one of the limits is infinity - isn't that called an improper integral of type 1 in English? So if that integral is supposed to converge, maybe there is something wrong with the formulas.

BTW, I just noticed that, while being good habit when working with vector parameter lists, it wouldn't even be necessary in case of those functions to vectorize.

A
AlanStevens19-Tanzanite
19-Tanzanite
May 18, 2013

Werner Exinger wrote:

But one of the limits is infinity - isn't that called an improper integral inEnglish?

I understand an improper integral to be one without any specific limits.

Alan

Terms & ConditionsCookie settingsAccessibility statement