Company
Mathematica Solutions to the ISSAC '97 Systems Challenge

Wolfram Research, Inc.


Problem 8

What is [Graphics:ISSACChallengegr142.gif]?

Result
[Graphics:ISSACChallengegr143.gif]


Method: Do a change of variables.

The current version cannot do the integral directly. (However, the next version can.)

[Graphics:ISSACChallengegr7.gif][Graphics:ISSACChallengegr144.gif]
[Graphics:ISSACChallengegr7.gif][Graphics:ISSACChallengegr145.gif]

In such a situation it is always a good idea to simplify the argument of the most complicated function. So we let [Graphics:ISSACChallengegr146.gif], and this works.

[Graphics:ISSACChallengegr7.gif][Graphics:ISSACChallengegr147.gif]
[Graphics:ISSACChallengegr7.gif][Graphics:ISSACChallengegr148.gif]

The substitution [Graphics:ISSACChallengegr149.gif] also works.

[Graphics:ISSACChallengegr7.gif][Graphics:ISSACChallengegr150.gif]
[Graphics:ISSACChallengegr7.gif][Graphics:ISSACChallengegr151.gif]

Note 1: We check the integral numerically to make sure the symbolic result is correct.

[Graphics:ISSACChallengegr7.gif][Graphics:ISSACChallengegr152.gif]
[Graphics:ISSACChallengegr7.gif][Graphics:ISSACChallengegr153.gif]

Note 2: The corresponding indefinite integrals can be done too.

[Graphics:ISSACChallengegr7.gif][Graphics:ISSACChallengegr154.gif]
[Graphics:ISSACChallengegr7.gif][Graphics:ISSACChallengegr155.gif]

[Graphics:ISSACChallengegr7.gif][Graphics:ISSACChallengegr156.gif]
[Graphics:ISSACChallengegr7.gif][Graphics:ISSACChallengegr157.gif]