Something wrong happening with definite integration with infinities. For example, integrate sin(x)/x from 0 to infinity, we get pi/2, but SMath get me answer, what looks like negative infinity. But if I integrate it from negative to positive infinity, it shows me that X is not defined. Same result happening when I integrate (x-1)/ln(x) and 1/x from 1 to infinity. Also, I checked integral from -1 to 1 of X, I got something looks like zero (2.9*10^-17 approximately zero), and integral from 0 to infinity of 1/x, which looks like infinity (3.3*10^11 approximately infinity for this software). Is it a bug? And will be it ever fixed? Which integration numerical methods using this software? And how can I calculate this integrals in correct way? Thank you for answering.

Wrote

Something wrong happening with definite integration with infinities. For example, integrate sin(x)/x from 0 to infinity, we get pi/2, but SMath does not integrate, what looks like negative infinity.

The Smath native Simpson integrator is often poor to scrap.
It does not like long range of the variable of integration.
The Romberg-Simpson kernel copes nicely for this last case.
For your other functions: please attach the Smath work sheet.

Jean

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Wrote

Something wrong happening with definite integration with infinities. For example, integrate sin(x)/x from 0 to infinity, we get pi/2, but SMath get me answer, what looks like negative infinity. But if I integrate it from negative to positive infinity, it shows me that X is not defined. Same result happening when I integrate (x-1)/ln(x) and 1/x from 1 to infinity. Also, I checked integral from -1 to 1 of X, I got something looks like zero (2.9*10^-17 approximately zero), and integral from 0 to infinity of 1/x, which looks like infinity (3.3*10^11 approximately infinity for this software). Is it a bug? And will be it ever fixed? Which integration numerical methods using this software? And how can I calculate this integrals in correct way? Thank you for answering.

Globally: your testing has no interest in Engineering works
x [-1 .. 1] = 0 by inspection
1/x = infinity by deduction
(x-1)/ln(x) [1 .. infinity] = infinity ... Non-sense coz no application.

(x-1)/ln(x) [0 .. some reasonable up] try Romberg-Simpson


When you are stuck, visit WolframAlpha on line integrator cost [0.00 $]
Always plot first.

@ 9838 km distance, my little eye can spy your red x not defined

This is my file, I wrote some comments here. I know about it's engineering application, and I know about wolfram alpha. But I search analogs for MathCAD, with client for Linux (and free will be good). I've found SMath, and my basic tasks here works great (studying tasks), but I've tried something, what I'm interested in, and this won't working correct in reason of radius of convergence of this numerical method. Thank you for answer. Also, your code for integration using Romberg Transform works fine for sin(x)/x, but gets something not correct for (x-1)/ln(x) from 0 to 1. Maybe reason is closest limits, does not working for this method.
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Will be back tomorrow, while you are up I'm horizontal in minutes.
Smath has been exhausted as much as Mathcad about numerical integration.
In short: Smath + utilities is equivalent to Mathcad.
The other method that does not lie and still the reference in many standards
is Legendre Quadrature. I have up to 63/64. They exists up to ~ 1000
but can't download ... see you soon.

Maple will do this, but Maxima failed in the recent nightly build.



Regards,
Radovan

Wrote

Maple will do this, but Maxima failed in the recent nightly build.



Regards,
Radovan

Thank you. Maple works great (but slowly, because running wine). But Maxima plugin doesn't work with Linux.

Wrote

But I search analogs for MathCAD, with client for Linux (and free will be good)

I have added few more of those Edu. Quiz
In short and within reason: Smath 1/1 Mathcad
Remember, no numerical maths are exact,
except arithmetic + ... accepted -,* sub-accepted /
NO function exist in computing machinery, all are approximations
In your Quiz and in maths what's the value ± ∞
Smath does not interpret more than what's computable.

Cheers ... Jean

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