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Positive infinity symbol mutates in a +/- infinity within integral (gamma function)
02-16-2014, 12:13 PM
Post: #1
Positive infinity symbol mutates in a +/- infinity within integral (gamma function)
I'm playing with CAS.

I tried to calculate the gamma function using my own definition (not the internal one).

g(x):=int(e^(-t)*t^(x-1),t,0,+infinity)

The CAS convert the +infinity to a +/-infinity. When you try to calculate the function using an integer value you obtain the right result:

g(5) gives 24

When you try to calculate the gamma for a rational number or a decimal you obtain this symbolic result:

g(5+1/3) gives int(t^5*e^(-t)/(t^(1/3))²,t,0,+/-infinity)

You can note that +infinity mutates in +/-infinity.

Why?

Massimo
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02-16-2014, 08:03 PM
Post: #2
RE: Positive infinity symbol mutates in a +/- infinity within integral (gamma function)
Doesn't happen to me.

define in CAS

g(t):=int(x^(t-1)*exp(-x),x,0,inf)

g(5)=24 OK

g(5+1/3)=int((3*x^4*x*e(-x))/3*(x^1/3)^(-2),x,0,inf) Note: No change to +/-inf

approx(Ans)=40.1289... OK
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02-16-2014, 09:43 PM
Post: #3
RE: Positive infinity symbol mutates in a +/- infinity within integral (gamma function)
(02-16-2014 08:03 PM)Helge Gabert Wrote:  Doesn't happen to me.

I have CAS in Exact mode.

Massimo
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02-17-2014, 12:22 AM
Post: #4
RE: Positive infinity symbol mutates in a +/- infinity within integral (gamma function)
So do I.
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02-17-2014, 06:28 AM (This post was last modified: 02-17-2014 06:38 AM by massimo.)
Post: #5
RE: Positive infinity symbol mutates in a +/- infinity within integral (gamma function)
On my calc or emulator, the problem appears when you set the Simplify settings in CAS to Minimum.
If you set it to None or Maximum the behavior is as expected.

BTW I set it to None and tried to obtain the same "error" by hand but I have not obtained any success so far.

Massimo
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02-17-2014, 10:53 AM
Post: #6
RE: Positive infinity symbol mutates in a +/- infinity within integral (gamma function)
The expand function changes +infinity in -/+infinity:

expand(int(t^x*e^(-t),t,0,+infinity))

gives

int(t^x*e^(-t),t,0,+/-infinity)

Massimo
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