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Wolfram Alpha bugs: How much time do you need to stamble over the 1st one?

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```It took me about 5 minutes.

Consider, as of  May 16, 2009, 8:45 a.m. local time (Simferopol)
Wolfram Alpha correctly returns 1 for the following integral

Integrate[BesselJ[1, z], {z, 0, Infinity}]

http://www.cybertester.com/Private/gs2.85/Wolfram_Alpha_bug-1-screen-1.png

But it leaves

Integrate[BesselJ[0, z], {z, 0, Infinity}]

unevaluated... while it is obviously = 1, too.

http://www.cybertester.com/Private/gs2.85/Wolfram_Alpha_bug-1-screen-2.png

I know almost for certain Robert Israel or someone
like Robert Israel will run to this place and say,
Oh c'mon, you nasty Bondarenko,
it's just a WEAKNESS not a bug..... ;)

Dammit, this looks like a common regression bug....

Mathematica 7 returns 1, but Wolfram Alpha fails...
```
 0
Reply vb914 (240) 5/16/2009 6:03:36 AM

```"Vladimir Bondarenko" <vb@cybertester.com> wrote in message
> It took me about 5 minutes.
>
> Consider, as of  May 16, 2009, 8:45 a.m. local time (Simferopol)
> Wolfram Alpha correctly returns 1 for the following integral
>
> Integrate[BesselJ[1, z], {z, 0, Infinity}]
>
> http://www.cybertester.com/Private/gs2.85/Wolfram_Alpha_bug-1-screen-1.png
>
> But it leaves
>
> Integrate[BesselJ[0, z], {z, 0, Infinity}]
>
> unevaluated... while it is obviously = 1, too.
>
> http://www.cybertester.com/Private/gs2.85/Wolfram_Alpha_bug-1-screen-2.png
>

The thing that I am not sure about, is what subset of Mathematica commands
does WA supports?

For if one enters say

Plot3D[Sin[x]*Cos[y], {x, -Pi, Pi},   {y, -Pi, Pi}]

WA will not process it, but if one enters

Plot[Sin[x]*Cos[x], {x, -Pi, Pi}]

it works.

So, clearly WA is not means to be the 'full' Mathematica kernel, else many
will no longer have a need to buy Mathematica itself if a free version is
on-line (even though it was a one command at a time).

So, I think the issue is that what you are using on WA web site is not the
same Mathematica one uses on the desktop, or there is some sort of filtering
out of commands that is being done, and may be the first command you used
was filtered out, just like plot3D was.  The filtering out might be based on
some criteria.

So, I do not think what you found is a 'bug', I think it is a matter of that
some Mathematica commands seem to be accepted, and some are not. But I could
be wrong of course.

But even with the subset of Mathematica commands WA seems to support, it is
still a nice thing to have available on-line to use when needed and for
free.

--Nasser

```
 0

```On May 16, 9:26=A0am, "Nasser Abbasi" <n...@12000.org> wrote:
> "Vladimir Bondarenko" <v...@cybertester.com> wrote in message
>
>
>
>
> > It took me about 5 minutes.
>
> > Consider, as of =A0May 16, 2009, 8:45 a.m. local time (Simferopol)
> > Wolfram Alpha correctly returns 1 for the following integral
>
> > Integrate[BesselJ[1, z], {z, 0, Infinity}]
>
> >http://www.cybertester.com/Private/gs2.85/Wolfram_Alpha_bug-1-screen-...
>
> > But it leaves
>
> > Integrate[BesselJ[0, z], {z, 0, Infinity}]
>
> > unevaluated... while it is obviously =3D 1, too.
>
> >http://www.cybertester.com/Private/gs2.85/Wolfram_Alpha_bug-1-screen-...
>
> The thing that I am not sure about, is what subset of Mathematica command=
s
> does WA supports?
>
> For if one enters say
>
> =A0 =A0Plot3D[Sin[x]*Cos[y], {x, -Pi, Pi}, =A0 {y, -Pi, Pi}]
>
> WA will not process it, but if one enters
>
> =A0 =A0 Plot[Sin[x]*Cos[x], {x, -Pi, Pi}]
>
> it works.
>
> So, clearly WA is not means to be the 'full' Mathematica kernel, else man=
y
> will no longer have a need to buy Mathematica itself if a free version is
> on-line (even though it was a one command at a time).
>
> So, I think the issue is that what you are using on WA web site is not th=
e
> same Mathematica one uses on the desktop, or there is some sort of filter=
ing
> out of commands that is being done, and may be the first command you used
> was filtered out, just like plot3D was. =A0The filtering out might be bas=
ed on
> some criteria.
>
> So, I do not think what you found is a 'bug', I think it is a matter of t=
hat
> some Mathematica commands seem to be accepted, and some are not. But I co=
uld
> be wrong of course.
>
> But even with the subset of Mathematica commands WA seems to support, it =
is
> still a nice thing to have available on-line to use when needed and for
> free.
>
> --Nasser

Hi Nasser,

I fully agree with you about Plot/Plot3D difference.

I'd find it too much funny if Wolfram Alpha can
handle OK

Integrate[BesselJ[1, z], {z, 0, Infinity}]
Integrate[BesselJ[3, z], {z, 0, Infinity}]
Integrate[BesselJ[5, z], {z, 0, Infinity}]

but fails here

Integrate[BesselJ[0, z], {z, 0, Infinity}]
Integrate[BesselJ[2, z], {z, 0, Infinity}]
Integrate[BesselJ[4, z], {z, 0, Infinity}]
Integrate[BesselJ[6, z], {z, 0, Infinity}]
Integrate[BesselJ[8, z], {z, 0, Infinity}]
Integrate[BesselJ[10, z], {z, 0, Infinity}]

and here

Integrate[BesselJ[7, z], {z, 0, Infinity}]
Integrate[BesselJ[9, z], {z, 0, Infinity}]
Integrate[BesselJ[11, z], {z, 0, Infinity}]

Is Wolfram Alpha a wizard the non-achiever?
```
 0
Reply vb914 (240) 5/16/2009 7:14:57 AM

```"Vladimir Bondarenko" <vb@cybertester.com> wrote in message
news:d5b8a711-cc7d-43bf-9750-

"I'd find it too much funny if Wolfram Alpha can
handle OK

Integrate[BesselJ[1, z], {z, 0, Infinity}]
Integrate[BesselJ[3, z], {z, 0, Infinity}]
Integrate[BesselJ[5, z], {z, 0, Infinity}]

but fails here

Integrate[BesselJ[0, z], {z, 0, Infinity}]
Integrate[BesselJ[2, z], {z, 0, Infinity}]
Integrate[BesselJ[4, z], {z, 0, Infinity}]
Integrate[BesselJ[6, z], {z, 0, Infinity}]
Integrate[BesselJ[8, z], {z, 0, Infinity}]
Integrate[BesselJ[10, z], {z, 0, Infinity}]
"

Yes. I am not sure what filtering is being done there.  Why it likes the odd
ones and not the even ones. You have a point here.

But for me, what is more important, is how to make WA understand more
general questions written in plain English.

I have been for the last hour trying to understand how to formulate good
questions to WA, but still having bit of trouble with some ones.

For example, how would you ask WA to answer the question if sin(x) is odd or
even function? or is it periodic?

I tried many ways, such as

is sin(x) an odd or an even function?
odd sin(x)
is sin(x) odd?
etc..

And
is sin(x) periodic?
etc..

But WA seems to have a bit of trouble with these questions, which I assume
because I am not formulating them in the way it could understand them. If
one however just types 'sin(x)', one gets lots of useful information about
the sin function.

--Nasser

```
 0

```NIntegrate[Sin[z^z], {z, 0, Infinity}]

-21.396

At any rate, the value of the integral is POSITIVE

http://www.cybertester.com/Private/gs2.85/Wolfram_Alpha_bug-2.pdf

Please note, NO warning message is generated...
```
 0
Reply vb914 (240) 5/16/2009 2:40:50 PM

```A finite value for a divergent integral.

NIntegrate[Sin[z], {z, 0, Infinity}]

1.

http://www.cybertester.com/Private/gs2.85/Wolfram_Alpha_bug-3.pdf

On May 16, 5:40=A0pm, Vladimir Bondarenko <v...@cybertester.com> wrote:
> NIntegrate[Sin[z^z], {z, 0, Infinity}]
>
> -21.396
>
> At any rate, the value of the integral is POSITIVE
>
> http://www.cybertester.com/Private/gs2.85/Wolfram_Alpha_bug-2.pdf
>
> Please note, NO warning message is generated...

```
 0
Reply vb914 (240) 5/16/2009 2:57:17 PM

```"Vladimir Bondarenko" <vb@cybertester.com> wrote in message
A finite value for a divergent integral.

NIntegrate[Sin[z], {z, 0, Infinity}]

1.

http://www.cybertester.com/Private/gs2.85/Wolfram_Alpha_bug-3.pdf

NIntegrate::"deodiv" : "\!\(\*
StyleBox[\"\"\", \"MT\"]\) DoubleExponentialOscillatory returns a finite \
integral estimate, but the integral might be divergent."

Uses the Mathematica Kernal after all.

```
 0
Reply VMCM1905 (13) 5/16/2009 4:05:40 PM

```Actually, the case we refer to in the 1st message

Integrate[BesselJ[0, z], {z, 0, Infinity}]

represents an intermittent bug.

Now you see the correct answer, 1, now you see
the unevaluated integral.

http://www.cybertester.com/Private/gs2.85/Wolfram_Alpha_bug-1-screen-2.png
http://www.cybertester.com/Private/gs2.85/Wolfram_Alpha_bug-1-screen-2b.pdf

On May 16, 9:03=A0am, Vladimir Bondarenko <v...@cybertester.com> wrote:
> It took me about 5 minutes.
>
> Consider, as of =A0May 16, 2009, 8:45 a.m. local time (Simferopol)
> Wolfram Alpha correctly returns 1 for the following integral
>
> Integrate[BesselJ[1, z], {z, 0, Infinity}]
>
> http://www.cybertester.com/Private/gs2.85/Wolfram_Alpha_bug-1-screen-...
>
> But it leaves
>
> Integrate[BesselJ[0, z], {z, 0, Infinity}]
>
> unevaluated... while it is obviously =3D 1, too.
>
> http://www.cybertester.com/Private/gs2.85/Wolfram_Alpha_bug-1-screen-...
>
> I know almost for certain Robert Israel or someone
> like Robert Israel will run to this place and say,
> Oh c'mon, you nasty Bondarenko,
> it's just a WEAKNESS not a bug..... ;)
>
> Dammit, this looks like a common regression bug....
>
> Mathematica 7 returns 1, but Wolfram Alpha fails...

```
 0
Reply vb914 (240) 5/16/2009 4:09:10 PM

```Yeah, but they wrongly defenestrate this CRITICALLY important message.

On May 16, 7:05=A0pm, "VMCM1905" <VMCM1...@gmail.com> wrote:
> "Vladimir Bondarenko" <v...@cybertester.com> wrote in message
>
> A finite value for a divergent integral.
>
> NIntegrate[Sin[z], {z, 0, Infinity}]
>
> 1.
>
> http://www.cybertester.com/Private/gs2.85/Wolfram_Alpha_bug-3.pdf
>
> NIntegrate::"deodiv" : "\!\(\*
> StyleBox[\"\"\", \"MT\"]\) DoubleExponentialOscillatory returns a finite =
\
> integral estimate, but the integral might be divergent."
>
> Uses the Mathematica Kernal after all.

```
 0
Reply vb914 (240) 5/16/2009 4:11:18 PM

```Nasser Abbasi schrieb:
> "Vladimir Bondarenko" <vb@cybertester.com> wrote in message
> news:d5b8a711-cc7d-43bf-9750-
>
> "I'd find it too much funny if Wolfram Alpha can
> handle OK
>
> Integrate[BesselJ[1, z], {z, 0, Infinity}]
> Integrate[BesselJ[3, z], {z, 0, Infinity}]
> Integrate[BesselJ[5, z], {z, 0, Infinity}]
>
> but fails here
>
> Integrate[BesselJ[0, z], {z, 0, Infinity}]
> Integrate[BesselJ[2, z], {z, 0, Infinity}]
> Integrate[BesselJ[4, z], {z, 0, Infinity}]
> Integrate[BesselJ[6, z], {z, 0, Infinity}]
> Integrate[BesselJ[8, z], {z, 0, Infinity}]
> Integrate[BesselJ[10, z], {z, 0, Infinity}]
> "
>
> Yes. I am not sure what filtering is being done there.  Why it likes the odd
> ones and not the even ones. You have a point here.
>

You are wondering how Wolfrom Research manage to protect the public
from coming to harm? Dear Gentlemen Detectives, this is most easily

First, they decided to switch off the unreliable Meijer-G machinery.

This leaves the table entry int x^(1-p) J_p(x) dx = -x^(1-p) J_(p-1)
(x) (cf. Gradshteyn-Ryzhik 5.52 2), and also the familiar recursion
relation for the Bessel functions (cf. Gradshteyn-Ryzhik 8.471 1)
which allows to lower the nu+p of x^nu J_p(x) in steps of two. As the
table entry covers the cases with nu+p = 1, this allows to integrate
all J_p(x) with odd integer order p>0, while clearly failing for even
orders.

In their highly responsible and commendable attitude, Wolfram Research
have further limited the recursion depth to prevent overheating and
eventual explosive disintegration of the kernel. You will easily see
how this explains the failure to integrate J_p(x) of order p=7 and
higher.

Hoping to having been of help,

Holmes.

```
 0
Reply clicliclic (36) 5/16/2009 4:14:01 PM

```Vladimir Bondarenko
Why is this here?  Doesn't Mathematica have its own newsgroup or support forum?
```
 0
Reply imageanalyst (7590) 5/16/2009 4:18:01 PM

```My dear Holmes, how on earth did you know that?! ;)

On May 16, 7:14=A0pm, cliclic...@freenet.de wrote:
> Nasser Abbasi schrieb:
>
>
>
> > "Vladimir Bondarenko" <v...@cybertester.com> wrote in message
> > news:d5b8a711-cc7d-43bf-9750-
>
> > "I'd find it too much funny if Wolfram Alpha can
> > handle OK
>
> > Integrate[BesselJ[1, z], {z, 0, Infinity}]
> > Integrate[BesselJ[3, z], {z, 0, Infinity}]
> > Integrate[BesselJ[5, z], {z, 0, Infinity}]
>
> > but fails here
>
> > Integrate[BesselJ[0, z], {z, 0, Infinity}]
> > Integrate[BesselJ[2, z], {z, 0, Infinity}]
> > Integrate[BesselJ[4, z], {z, 0, Infinity}]
> > Integrate[BesselJ[6, z], {z, 0, Infinity}]
> > Integrate[BesselJ[8, z], {z, 0, Infinity}]
> > Integrate[BesselJ[10, z], {z, 0, Infinity}]
> > "
>
> > Yes. I am not sure what filtering is being done there. =A0Why it likes =
the odd
> > ones and not the even ones. You have a point here.
>
> You are wondering how Wolfrom Research manage to protect the public
> from coming to harm? Dear Gentlemen Detectives, this is most easily
>
> First, they decided to switch off the unreliable Meijer-G machinery.
>
> This leaves the table entry int x^(1-p) J_p(x) dx =3D -x^(1-p) J_(p-1)
> (x) (cf. Gradshteyn-Ryzhik 5.52 2), and also the familiar recursion
> relation for the Bessel functions (cf. Gradshteyn-Ryzhik 8.471 1)
> which allows to lower the nu+p of x^nu J_p(x) in steps of two. As the
> table entry covers the cases with nu+p =3D 1, this allows to integrate
> all J_p(x) with odd integer order p>0, while clearly failing for even
> orders.
>
> In their highly responsible and commendable attitude, Wolfram Research
> have further limited the recursion depth to prevent overheating and
> eventual explosive disintegration of the kernel. You will easily see
> how this explains the failure to integrate J_p(x) of order p=3D7 and
> higher.
>
> Hoping to having been of help,
>
> Holmes.

```
 0
Reply vb914 (240) 5/16/2009 5:03:47 PM

```"Image Analyst" <imageanalyst@mailinator.com> wrote in message <gumovp\$ib4\$1@fred.mathworks.com>...
> Why is this here?  Doesn't Mathematica have its own newsgroup or support forum?

It does. As usual for him, this is off topic spam.

John
```
 0
Reply woodchips (7921) 5/16/2009 5:21:01 PM

```Yet another bug.

If you try to calculate the integrals like this one

Integrate[SinIntegral[z]^2/z^2,{z,0,Infinity}]

you see no answer at the page.

```
 0
Reply vb914 (240) 5/17/2009 10:14:18 PM

```Vladimir Bondarenko <vb@cybertester.com> wrote in message <5a351224-1c4c-4f44-98bf-587c8b409c78@g20g2000vba.googlegroups.com>...
> Yet another bug.
>
> If you try to calculate the integrals like this one
>
> Integrate[SinIntegral[z]^2/z^2,{z,0,Infinity}]
>
> you see no answer at the page.
>
> see the correct answer, Pi.

stop spamming this NG (CSSM), which deals with matters around MATLAB...

us
```
 0
Reply us1 (8054) 5/17/2009 10:52:01 PM

```Vladimir Bondarenko wrote:
> Yet another bug.
>
> If you try to calculate the integrals like this one
>
> Integrate[SinIntegral[z]^2/z^2,{z,0,Infinity}]
>
> you see no answer at the page.

I'm surprised Wolfram Research don't put a note such as "For this, you
will need to purchase Mathematica"

>
> see the correct answer, Pi.

Does that work with all Mathematica expressions? If so, I assume WRI
will wish to fix this, as they are unlikely to want to make the full
functionality of Mathematica available over the web.

I just found an interesting one with PrimePi on Wolfram Alpha. The
results from that appear to be one of 3 forms.

* PrimePi[n] returns a result if n is small (say 2^10).

* PrimePi[n] returns nothing useful at all if n is large (say 2^40), but
computable by Mathematica (the answer is 41203088796). Clearly WRI don't
want to spend too much CPU time computing things. I've no idea where it
stops working on Wolfram Alpha.

* PrimePi[n] gives a lot of information about primes if n is greater
than 249999999999999, which can't be computed by Mathematica at all (it
gives up for numbers greater than 249999999999999. I've no idea why WRI
have that limit.

BTW Vladimir, Wolfram Alpha has only just been released, so it's no
surprise there are bugs. I hope you don't get into the habit of posting
endless bugs of Wolfram Alpha to countless newsgroups. It would be
better to send them directly to WRI via the link on the page.

It would be even better if you could use your bug-hunting skills to
something like Sage and provide the Sage developers with useful feedback
on any bugs you find.

--
I respectfully request that this message is not archived by companies as
unscrupulous as 'Experts Exchange' . In case you are unaware,
'Experts Exchange'  take questions posted on the web and try to find
idiots stupid enough to pay for the answers, which were posted freely
by others. They are leeches.
```
 0
Reply foo25 (218) 5/18/2009 10:46:13 AM

```Vladimir Bondarenko wrote:
> Yet another bug.
>
> If you try to calculate the integrals like this one
>
> Integrate[SinIntegral[z]^2/z^2,{z,0,Infinity}]
>
> you see no answer at the page.
>
> see the correct answer, Pi.

I can't really imagine most people care to run integrals through Wolfram
Alpha, though it is nice that it does some of them. Clearly the Google
calculator could be hooked up to a computer algebra system, thereby
making such questions "easy" to answer, subject to the availability of
computer time. Integration is hardly a major resource for most people.

WA is supposed to know about People & History.

I tried "when did abraham lincoln die"  and WA got april 15, 1865.
but "who died on april 15, 1865"  got "WA isn't sure what to do with
Nor did "what happened on april 15, 1865" get any results.
"die april 15" returns with a suggestion that it interpreted "die" as
"Die, Rhone-Alpes, France".

I think that if you want to find weaknesses in WA, looking for obscure
integrals is hardly the primary spot.  But then sci.math.symbolic is not
the right place to discuss this.

```
 0
Reply fateman2 (331) 5/18/2009 1:55:24 PM

```One interesting mistake. Using Wolfram|Alpha one can find that:
sin'(2x) = cos(2x)

:-)))
```
 0

```Vladimir MarjanoviÄ‡ wrote:
> One interesting mistake. Using Wolfram|Alpha one can find that:
> sin'(2x) = cos(2x)
>
> :-)))

Not a mistake!  Putting a prime on the name of a function
commonly denotes the corresponding derivative function, not the
result of applying the derivative operator to the function
evaluated at another function.

In this case, sin'(x) = cos(x) for all x implies sin'(ax) =
cos(ax) for all a.

-- David
```
 0
Reply williams1 (514) 5/18/2009 5:39:59 PM

```["Followup-To:" header set to sci.math.]

On 2009-05-18, Vladimir Marjanovi? <nemam.email@joj.mene.hr> wrote:
> One interesting mistake. Using Wolfram|Alpha one can find that:
> sin'(2x) = cos(2x)

That's not a bug, is it?  If I have f'(x) = g(x), then I'd certainly
expect that f'(2x) = g(2x).  The prime denotes the derivative of f
with respect to its argument -- it doesn't magically differentiate the
entire expression with respect to x.  If you want df(2x)/dx, just say
so; don't abuse a well-established notation that means something else.

--
Ilmari Karonen
```
 0
Reply usenet210 (9) 5/18/2009 5:46:18 PM

```Well, it turned out that it's easy to find a bug in the Wolfram Alpha.
Also, now it's easy to make the proper screen shots.

Here you are.

Say, Wolfram Alpha returns for this simple integral

Integrate[Sin[z]^2/Sinh[z]^2, {z, 0, Infinity}]

the following value

(2*Pi*Coth[Pi]-1)/4 = 1.3266...

http://www.cybertester.com/Private/gs2.85/Wolfram_Alpha_bug-4-symbolic.pdf

which is obviously wrong as

NIntegrate[Sin[z]^2/Sinh[z]^2, {z, 0, Infinity}]

1.07667

The correct integral value is

(Pi*Coth[Pi]-1)/2 = 1.076674047468581174...

;)

http://www.cybertester.com/Private/gs2.85/DSCN0003.JPG
http://www.cybertester.com/Private/gs2.85/DSCN0037h.JPG
http://www.cybertester.com/Private/gs2.85/DSCN0058.JPG

```
 0
Reply vb914 (240) 5/19/2009 6:37:35 PM

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