Why Does int() of rectangularPulse Return NaN?

syms t real
x(t) = rectangularPulse(0,1,t);
int(x(t),t,0,5)
ans = 
1
int(x(t),t,0,inf)
ans = 
NaN
int(x(t),t,-10,10)
ans = 
1
int(x(t),t,-inf,inf)
ans = 
NaN
Any ideas why those two cases return NaN?

1 Kommentar

Fixed in 2022a
syms t real
x(t) = rectangularPulse(0,1,t);
int(x(t),t,0,inf)
ans = 
1
int(x(t),t,-inf,inf)
ans = 
1
int(rectangularPulse(0,1,t),-2,inf)
ans = 
1

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 Akzeptierte Antwort

Work-around:
syms b x t real
assume(b>=0)
y(t) = rectangularPulse(x,1,t)
y(t) = 
z = int(y,t,0,b)
z = 
limit(z, b, inf)
ans = 

4 Kommentare

Paul
Paul am 9 Jul. 2021
Seems like this should be pretty simple function for int() to deal with. Does this seem like a bug?
Walter Roberson
Walter Roberson am 10 Jul. 2021
Bearbeitet: Walter Roberson am 10 Jul. 2021
I am not sure why this is happening. You can read the internal code for rectangularPulse by using
regexprep(char(feval(symengine, 'expose', 'rectangularPulse')),'\\n','\n')
but that doesn't tell you anything about how it integrates. The symbolic integration routine is too large for me to chase through.
Paul
Paul am 10 Jul. 2021
Bearbeitet: Paul am 10 Jul. 2021
Just seems so strange because int() handles much more complex functions, which is just about any function, with ease. I'll see what Tech Support says about this.
Another interesting result:
syms t real
int(rectangularPulse(0,1,t),-inf,2)
ans = 
1
int(rectangularPulse(0,1,t),-2,inf)
ans = 
NaN
I have a suspicion that somewhere along the way, a dirac(0) is getting invoked.

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Version

R2021a

Gefragt:

am 9 Jul. 2021

Kommentiert:

am 25 Mai 2022

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