Wikipedia:Reference desk/Archives/Mathematics/2016 November 30

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November 30

Wolfram Alpha indefinite integral

According to Wolfram Alpha, $\int x^{\sqrt {x}}dx={\frac {x^{{\sqrt {x}}+1}}{{\sqrt {x}}+1}}+C$. Numerically and algebraically this doesn't seem right (I don't think the differential of the right hand side simplifies to the left). Does Mathematica also give this result? If this is wrong, is there a nice closed form, or what is WA doing here? 24.255.17.182 (talk) 04:58, 30 November 2016 (UTC)

• Convenience links: [1], [2]. It does not look right, but I have no idea what is happening. Tigraan^{Click here to contact me} 08:14, 30 November 2016 (UTC)
• What it's doing is clearly applying the standard rule for integrating polynomials ($\int x^{a}dx={\frac {x^{a+1}}{a+1}}+C$) and then substituting $a={\sqrt {x}}$ without realizing that this is itself a function of x. This is presumably wrong. As far as I can tell, there is no closed form solution. Otherwise, by making the substitution $y={\sqrt {x}}$, you could rewrite it as $\int 2y^{3y}dy$, which when you ignore the constants is effectively $\int y^{y}dy$, and this definitely has no closed solution. Smurrayinchester 09:35, 30 November 2016 (UTC)

Mathematica does not have this error. Evaluating Integrate[x^Sqrt[x], x] simply gives no results, which strongly indicates that there is no elementary solution. And yes, it does seem that in this case WA treats the ${\sqrt {x}}$ as a constant even though it's obviously not.

I have submitted a bug report to Wolfram about this. I wouldn't keep my hopes up though, there is a different bug report I've submitted a year ago which still hasn't been resolved.

I'm a big fan of Mathematica but things like this are rather disappointing. -- Meni Rosenfeld (talk) 09:58, 30 November 2016 (UTC)

So I haven't seen an actual refutation yet. The tempting error looks suspect, but there are things like that — see sophomore's dream.

But if I haven't made a mistake, it's not hard to refute numerically. Letting

$z={\frac {x^{{\sqrt {x}}+1}}{{\sqrt {x}}+1}}$

then applying logarithmic differentiation, we get

${\frac {1}{z}}{\frac {dz}{dx}}={\frac {{\sqrt {x}}+1}{x}}+{\frac {1}{2{\sqrt {x}}}}\log x-{\frac {1}{2{\sqrt {x}}({\sqrt {x}}+1)}}$

${\frac {dz}{dx}}=z*\mathrm {above\ quantity}$

which I plugged into a spreadsheet to compare with the integrand, and they are not the same. For example, for x=0.0001, I get dz/dx=0.8659568231, but the integrand is 0.9120108394. I could have made a mistake of course. Someone could check me. --Trovatore (talk) 21:07, 30 November 2016 (UTC)

Numeric refutation is the most direct, but is not easy to share. I did my own numeric evaluation before moving further, but it's not like I was going to actually post the entire calculation here. Perhaps I should have written something to the effect of "I have refuted it numerically", but it seemed too obvious.

Furthermore, there were actually two refutations here: The link by Tigraan that shows the derivative of the result is quite different from the original function; and Smurrayinchester's proof that if the function did have an elementary antiderivative, a different function would also have one which I believe is known to have none. -- Meni Rosenfeld (talk) 23:01, 30 November 2016 (UTC)

Well, checking whether two complicated expressions are equal is a hard problem, I believe undecidable in general. And even if we know that there is no elementary antiderivative of x^x, it's not immediate that there's none for x^3x. --Trovatore (talk) 23:34, 30 November 2016 (UTC)