Wolfram Alpha numerical errors, integral computation,

Question

Just out of curiosity, I wonder why Wolfram Alpha would give me two different answers to the same query that very much do not agree. Numerical computation of an integral is involved (with big powers of small $x$'s, so big rounding off errors, enlarged by a factor of $2^{1000}$), and I have no reason to expect that the error in either of the two answers is small (so neither of the two approximations need to be close to the correct value), but I wonder what are the inner workings of Wolfram Alpha that make it display two answers that are so off from each other.

One answer is $1.6036\times 10^{105}$, the other is
$-2.041253178575746070624332792862500105708912774979593... \times 10^{100}$.

Note the opposite signs, and the big absolute values, so anything could be in between.

I just noticed Wolfram shows some more details, the first answer is "Computation result:"

2^1000 integral_(1/2)^(2/3) x^1000 sin(1/(2/3 - x)^2) dx = 1.6036×10^105

and the second answer is "Decimal approximation:"

-2.041253178575746070624332792862500105708912774979593... × 10^100

where "Wolfram Language code:"

N[2^1000 Integrate[x^1000 Sin[1/(2/3 - x)^2], {x, 1/2, 2/3}], 79]

The query is
((2^1000)* (integral from 1/2 to 2/3 of x^1000 sin(1/(2/3-x)^2)))

i.e. $2^{1000}\int_{1/2}^{2/3}x^{1000}\sin(\frac1{(\frac23-x)^2})dx$

link to Wolfram Alpha query

I was trying to get a better understanding of some integrals related to the following MSE problem.

Answer 1

This is not a complete answer, but for what it's worth, trying your code

N[2^1000 Integrate[x^1000 Sin[1/(2/3 - x)^2], {x, 1/2, 2/3}], 79]

in Mathematica gives

-2.041253178575746070624332792862500105708912774979593349983494809132420022371226*10^100

According to the documentation, “N[Integrate[…]] calls NIntegrate for integrals that cannot be done symbolically”, with the option PrecisionGoal by default being 10 less than WorkingPrecision. And the call

NIntegrate[2^1000 x^1000 Sin[1/(2/3 - x)^2], {x, 1/2, 2/3}, WorkingPrecision -> 89]

indeed returns

 -2.0412531785757460706243327928625001057089127749795933499834948091324200223712255170998619*10^100

(but much faster, probably because it doesn't try symbolic integration first).

So that seems to be the source of the negative answer. Where WA gets that positive answer, I have no clue!

Answer 2

This is just a comment about how to find the correct numerical value of the integral. Taking $x = 2/3 - 1/\sqrt z$, we obtain $$I = \int_{1/2}^{2/3} x^\lambda \sin \frac 1 {\left( \frac 2 3 - x \right)^2} dx = \int_{36}^\infty \frac {x^\lambda \sin z} {2 z^{3/2}} dz = \operatorname {Im} \int_{36}^\infty \frac {e^{\lambda \ln x + i z}} {2 z^{3/2}} dz.$$ The relevant saddle point of the exponent is the root $z_0$ of $$16 z^3 - 36 z^2 - 36 i \lambda z + 9 \lambda^2$$ located in the first quadrant. The ray $\gamma$ from $36$ through $z_0$ does not go through $z_0$ in the steepest descent direction, but it is good enough, as the integrand is no longer oscillatory:

Thus numerical integration along $\gamma$ is straightforward and doesn't require bignum arithmetic. Taking $\lambda = 10^3$ gives $$2^\lambda I = 2^\lambda \operatorname {Im} \int_\gamma \frac {e^{\lambda \ln x + i z}} {2 z^{3/2}} dz \approx -6.959 \times 10^{26}.$$