Why are the values of z1 and z2 false in the special case a=0?

Hi,

Why are the values of z1 and z2 false in the special case a=0?

clear
syms  t
syms a real
x(t) =exp(-a*t)*heaviside(t)

x(t) = 

z1 = int(x(t),t, -inf, inf)

z1 = 

z2 = int(exp(-a*t),t,0, +inf)

z2 = 

A big thank you in advance!

[Torsten]

Why are the values of z1 and z2 false in the special case a=0?

Because the Symbolic Toolbox has a bug. You should report your observation to Technical Support.

[David Goodmanson]

Hi Mohamed,

What would you prefer the result to be if a = 0? The present result,

1/a for 0 <= a

certainly is unbounded as a --> 0 from above (and coming from above is implied by the <= sign). Effectively a is infinite at a = 0, which is the correct result.

[Torsten]

Mathematica gives for the first integration and a = 0:

Integral does not converge.

And for the second integration the case a=0 should definitly be included in the second case result oo.

[Walter Roberson]

Testing what it should be if a is 0, which gives exp(-0*t) which gives exp(0) which is 1

syms t
x(t) = heaviside(t);
z1 = int(x, t, -inf, +inf)

z1 = 

∞

z2 = int(x, t, 0, +inf)

z2 = 

∞

[Paul]

Trying it this way does not even allow for the possibility that a = 0.

syms  t
syms a real
x(t) = exp(-a*t)*heaviside(t);
syms b n1 n2
assume([b n1 n2] < 0)
syms c p1 p2 positive

z1 = limit(int(x(t),t,b,n1),b,-inf) + limit(int(x(t),t,n1,n2),n2,0,'Left')  ...
    + limit(int(x(t),c,p1),c,0,'Right') + limit(int(x(t),p1,p2),p2,inf);
z1 = simplify(z1)

z1 = 

z2 = limit(int(exp(-a*t),t,c,p1),c,0,'Left') + limit(int(exp(-a*t),t,p1,p2),p2,inf);
z2 = simplify(z2)

z2 = 

subs(z2,a,0)

ans = 

NaN

Apparently int evaluates these imporoper integrals with a different approach, though I thought what I did here is the formal defintions of these improper integrals.

I'm actually quite surprised by the results in the original question insofar as I don't see int return a piecewise result for other cases where it could.

For example:

f(t) = triangularPulse(t);
syms n integer
c(n) = int(exp(-1j*2*sym(pi)/4*n*t)*f(t),t,-inf,inf)

c(n) = 

try
    c(0);
catch ME
    ME.message
end

ans = 'Division by zero.'

However, c(0) is well defined, as we can see by subbing in n = 0 into the integrand and then integrating

C0 = int(exp(-1j*2*sym(pi)/4*0*t)*f(t),t,-inf,inf)

C0 = 

1

Why didn't int() return a piecewise expression for c(n) in the first place? I've always thought that "int just doesn't do that," but this Question shows that it can. I wonder if int recognizes the expression for c(n) as a Fourier integral and resorts to fourier instead of computing the integral other using other means.

Answer 1

[SAI SRUJAN]

Hi Mohamed,

I understand that you are trying to examine the values of 'z1' and 'z2' in a special case where 'a = 0'.

In this specific case, both 'z1' and 'z2' approach infinity, because the integrals of the functions 'exp(-a*t)' and 'exp(-a*t)*heaviside(t)' become improper integrals that diverge when 'a = 0'. Let's break down each integral:

For `z1`, the integral is calculated as:

z1 = int(x(t), t, -inf, inf)

Since `a = 0`, the exponential term becomes `exp(0*t) = 1`, and the integral becomes:

z1 = int(1, t, -inf, inf)

This is the integral of a constant function over an infinite interval, which is unbounded and thus diverges to infinity.

For `z2`, the integral is calculated as:

z2 = int(exp(-a*t), t, 0, +inf);

Since `a = 0`, the exponential term becomes `exp(0*t) = 1`, and the integral becomes:

z2 = int(1, t, 0, +inf)

This integral also represents an area under the constant function 1 from 0 to inf, which is again unbounded.

In the particular situation where 'a = 0', the integrals for both 'z1' and 'z2' turn out to be infinite. Therefore, they are deemed "inf" since the result of the integration isn't a finite value.

I hope this helps!