vladimir bondarenko

Hello the inspired formula lovers,

It's refreshing to observe how human mind fights against
dark of ignorance, against complexity trying to discover
patterns, - and enjoy the victories gained!

None of modern computer algebra systems can crack this
integral straightforwardly. Such a sad fact...

Is there a soul who can show a way to calculate using a
CAS the exact value of the integral

int(cos(z^2)*sech(z*sqrt(Pi))^3, z= 0..infinity);

?


Best wishes,

Vladimir Bondarenko

VM and GEMM architect
Co-founder, CEO, Mathematical Director

http://www.cybertester.com/ Cyber Tester, LLC
http://maple.bug-list.org/ Maple Bugs Encyclopaedia
http://www.CAS-testing.org/ CAS Testing

/ 1/2 1 \ 1/2
|Pi - -----| 2
| 1/2|
\ Pi /
-------------------- 4


0.4271859285

Enjoy!

Igor

tcl

My Maple 9 cannot get this. Pretty sad.

vladimir bondarenko

TCL writes:

TCL> My Maple 9 cannot get this. Pretty sad.


Maple 9 ?

If you believe that ANY Maple version can calculate
this integral

int(cos(z^2)*sech(z*sqrt(Pi))^3, z= 0..infinity);

then... Pretty sad

vladimir bondarenko

Dau, you again as usually meet the challenge

I know how I did it; but how you did, pleeease? ;-)

('pon my word, I like so much your explanations!)


Vladimir

g. a. edgar

Maple 9.5 cannot either, but itself. Igor gave it some help.

--
G. A. Edgar http://www.math.ohio-state.edu/~edgar/

igor khavkine

Vladimir, I can't be exactly sure how you composed this integral, but
the sqrt(Pi) facter was a dead give away. :-)

Here's the help that I had to give Maple for this integral:

# The trick is to extend the integration to z=-infinity and to shift
# the integration contour by I*p. The integral can then be evaluated
# in terms of the residues of the integrand, inside the box bounded by
# the original and shifted contours, and the auxiliary integral
#
# Int(cos(z^2)*sech(z*p), z=-infinity..infinity);
#
# the latter can be evaluated using the same trick but on the integral # Int(cos(z^2)*sech(z*p)*csch(z*p)^2, z=-infinity..infinity);


1/2 p := Pi

2 1/2 3
f := cos(z ) sech(z Pi )

2 1/2 1/2 2
g := cos(z ) sech(z Pi ) csch(z Pi )


1/2
1/2 I 2
R12 := ----------
3/2 Pi

bytes used=164061396, alloc=5897160, time=7.80
1/2
1/2 I 2
Rp12 := ----------
1/2
Pi

# The residues of g at integral multiples or I*pi are all zero
# This simplified gives the answer I first posted.


/ 1/2 1/2\
|1/2 I 2 1/2 I 2 |
myJ := 1/2 I Pi |---------- - ----------|
| 3/2 1/2 |
\ Pi Pi /


0.427185928457033786633968291235570379013817369570 00


myI := 0.427185928457033786633968291235570379013817369570 02

Igor