Wolfram Alpha (free) Limitations

[PeroK]

When I put the following integral into the free version of Wolfram I get a sensible answer:
$$\int \frac{1}{(1 - \frac 1 u)\sqrt{\frac 1 u - a}}du$$The function includes an inverse $\tanh$ function that (as expected) diverges as $u \to 1$. But, if I try:
$$\int \frac{1}{(1 - \frac 1 u)\sqrt{\frac 1 u - \frac 1 a}}du$$I get a much more complicated result without the inverse $\tanh$, which appears not to diverge as $u \to 1$.

The moral is, I guess, even simplifying a single parameter can make things easier for the engine.

 

[Wrichik Basu]

Interesting. For comparison, I carried out the integral in MATLAB using the Symbolic Math Toolbox. Here is the code I used:

syms u a
ex1 = 1 ./ ((1 - 1/u) .* sqrt(1/u - a));
i1 = int(ex1, u);

ex2 = 1 ./ ((1 - 1/u) .* sqrt(1/u - 1/a));
i2 = int(ex2, u);

u = 1;

simplify(subs(i1))
simplify(subs(i2))

Line 10 returned NaN.

Line 11 returned the following expression:
$$\sqrt{-a}\,\mathrm{atanh}\left(\sqrt{-a}\,\sqrt{\frac{a-1}{a}}\right)\,\left(a+2\right)-a\,\sqrt{\frac{a-1}{a}}+\frac{\mathrm{atan}\left(\frac{{\left(a\,\left(a-1\right)\right)}^{7/2}\,\sqrt{\frac{a-1}{a}}\,1{}\mathrm{i}}{a^3\,{\left(a-1\right)}^4}\right)\,\sqrt{a\,\left(a-1\right)}\,2{}\mathrm{i}}{a-1}$$
It still has $\mathrm{atanh}$, but is not NaN.

 

[PeroK]

 

[vela]

Mathematica gives the following result for the original integral:
$$-\frac{(2 a+1) \tan ^{-1}\left(\frac{\sqrt{\frac{1}{u}-a}}{\sqrt{a}}\right)}{a^{3/2}}-\frac{u \sqrt{\frac{1}{u}-a}}{a}+\frac{2 \tan ^{-1}\left(\frac{\sqrt{\frac{1}{u}-a}}{\sqrt{a-1}}\right)}{\sqrt{a-1}}$$ For the second integral, it gives the same answer with $a$ replaced by $1/a$.

 

[PeroK]

Mathematica gives the following result for the original integral:
$$-\frac{(2 a+1) \tan ^{-1}\left(\frac{\sqrt{\frac{1}{u}-a}}{\sqrt{a}}\right)}{a^{3/2}}-\frac{u \sqrt{\frac{1}{u}-a}}{a}+\frac{2 \tan ^{-1}\left(\frac{\sqrt{\frac{1}{u}-a}}{\sqrt{a-1}}\right)}{\sqrt{a-1}}$$ For the second integral, it gives the same answer with $a$ replaced by $1/a$.

Suprisingly the derivatives of those functions are the same:
$$\frac{d}{du}\bigg (\frac{2 \tan ^{-1}\left(\frac{\sqrt{\frac{1}{u}-a}}{\sqrt{a-1}}\right)}{\sqrt{a-1}} \bigg ) = \frac{d}{du}\bigg (-\frac{2 \tanh ^{-1}\left(\frac{\sqrt{\frac{1}{u}-a}}{\sqrt{1-a}}\right)}{\sqrt{1-a}} \bigg )$$
When Wolfram produced the version with the second $\tan^{-1}$ function, I thought it must be wrong, as I needed the $\tanh^{-1}$ to convert to a log function.

 

[vela]

It took me awhile to figure out how it diverged. When $u=1$, the argument of the second arctan is equal to $i$, and apparently, arctan has a pole there. I never knew that.

 

[PeroK]

Suprisingly the derivatives of those functions are the same:
$$\frac{d}{du}\bigg (\frac{2 \tan ^{-1}\left(\frac{\sqrt{\frac{1}{u}-a}}{\sqrt{a-1}}\right)}{\sqrt{a-1}} \bigg ) = \frac{d}{du}\bigg (-\frac{2 \tanh ^{-1}\left(\frac{\sqrt{\frac{1}{u}-a}}{\sqrt{1-a}}\right)}{\sqrt{1-a}} \bigg )$$
When Wolfram produced the version with the second $\tan^{-1}$ function, I thought it must be wrong, as I needed the $\tanh^{-1}$ to convert to a log function.

Of course, the functions apply to different regions. The LHS is valid for $a > 1$ and the RHS for $a < 1$.

 

[PAllen]

I once had a correspondence with Wolfram about their integrator. I had an integral that it wouldn't solve at all that I thought "ought" to be solvable. Similar to the OP case, I made the most obvious variable substitution, which made it no easier for me, but then Wolfram solved it no problem. Wolfram actually claimed that they didn't consider this a bug, and that the same would have happened on the fully paid version. It may have changed since, but they said Mathematica and Alpha do not do substitutions per se, and there would always be cases like the one I ran into.

I am also bothered by a tendency of the free version to unnecessarily include complex functions in their solutions for real integrals.

The upshot is you should only use it as an assist, combined with standard hand methods, which is fine. I am not normally looking for a black box to do everything.