Simple problems that Wolfram can't do

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Steve Zissou
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Hi everyone, I thought I would share a thought - apparently simple problems that Wolfram can't seem to handle. I'll go first:
$$ \int_{0}^{\infty}e^{-x^p}dx=\frac{1}{p}\Gamma\left( \frac{1}{p} \right) $$
Entering this into Wolfram alpha:
Integrate[Exp[-x^p],{x,0,\infty}]
Gets you nowhere. Ha!
Please share your own frustrations also. Thanks!
 
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But Wolfram Mathematica has no problem:
1746389059318.webp

I guess Alpha is a stripped-down Mathematica Lite!
 
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renormalize said:
But Wolfram Mathematica has no problem:
View attachment 360674
I guess Alpha is a stripped-down Mathematica Lite!
Well that's weird.
 
https://www.integral-calculator.com/ works.

Happily brought to you by our useful list of calculators:

1746390294240.webp
 
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Steve Zissou said:
Hi everyone, I thought I would share a thought - apparently simple problems that Wolfram can't seem to handle. I'll go first:
$$ \int_{0}^{\infty}e^{-x^p}dx=\frac{1}{p}\Gamma\left( \frac{1}{p} \right) $$
Entering this into Wolfram alpha:
Integrate[Exp[-x^p],{x,0,\infty}]
Gets you nowhere. Ha!
Please share your own frustrations also. Thanks!
I got Wolfram Alpha to give me the correct answer by submitting the problem again right after it failed the first time. I think you're just running into resource limits of the free version.
 
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FWIW ChatGPT got it

1746474166364.webp
 
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vela said:
I got Wolfram Alpha to give me the correct answer by submitting the problem again right after it failed the first time. I think you're just running into resource limits of the free version.
@vela Whoa that's unexpected. So it will do it, but there is some resource limits that I was unaware of?
 
BWV said:
FWIW ChatGPT got it

View attachment 360706
@BWV Ok that's unexpected. I've had AI give some wacky math answers so I haven't tried using that tech for a while for math.
 
pines-demon said:
WolframAlpha cannot do simple contour integration. Can it?
I'm not sure how to tell WolframAlpha if certain variables are complex, real, or any sort of bounds on parameters.
 
Ok I kid you not on this one:
Integrate[t^(z-1)*Exp[-t],{t,0,\infty}]
Yields the answer:

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