Improper (difficult) integrals at SwissMicros Calculator Forum

[peacecalc]

Hello Valentin,

thank you very much for your quick and comprehensive answer. I didn't expect
this. I feel esteemed.

off topic:
I regret and feel sorry, that you had made bad experiences with some
members of this forum. Using fora is a tricky thing. As a member
in some other fora I saw that even one forum died because of a bad ping-pong
post game between only few members.

ontopic:
About your question:
I used the native programming language of the hp prime (PPL).
Maybe I give the builtin Python a chance, but for getting a
higher precision the number of calls have to raise with factor 10,
that means, instead of 4, I have to wait 40 minutes.
So Python has to be much more faster then PPL.

I carried out the estimated Integral:

\[ I = \int_{-1}^{1}\exp\left(x\cdot\left(\frac{1+\sqrt{5}}{2}\right)\right)dx = (\sqrt{5}-1)*\sinh\left(\frac{\sqrt{5}+1}{2}\right) \approx 2.9943\]

thats not bad for such an easy estimation.

Because there only a few valid digits,
it is clear, that I haven't found a secrect relation, it is only
a not bad guess.

@Ajaja: Just carry out the integral

\[ s = \int_{-1}^{1}\sqrt{\left(1+f'(x)^2\right)}dx =
\int_{-1}^{1} \sqrt{\left(1+\left((1+e^{x+\frac{1}{3}} e^{e^{x+\frac{1}{3}}} e^{e^{e^{x+\frac{1}{3}}}}\cos\left(e^{e^{e^{x+\frac{1}{3}}}}\right)) e^{x+\sin\left(e^{e^{e^{x+\frac{1}{3}}}}\right)} \right)^2\right)}dx \]

Maybe a substitution makes it easier..., but it looks ugly, very ugly...
By the way, for writing down the integral in my poste, I used the command "latex" of hp prime. I try to make it smoother...

\[ \int_{-1}^{1}\sqrt{1+f'(x)^2}dx =
\int_{-1}^{1} \sqrt{1+\left(\left(1+ \exp\left(x+\frac{1}{3}\right) \exp\left(\exp\left(x+\frac{1}{3}\right)\right)\exp\left(\exp\left(\exp\left(x+ \frac{1}{3}\right)\right)\right)
\cos\left(\exp\left(\exp\left(\exp\left(x+ \frac{1}{3}\right)\right)\right)\right) \right)
\exp\left(x+\sin\left(\exp\left(\exp\left(\exp\left(x+\frac{1}{3}
\right)\right)\right)\right)\right)\right)^2}dx \]

However I do, the integral is still ugly. Editing was necessary because of false brackets setting and typo at estimated value.