Post Reply 
Quiz: calculating a definite integral
01-03-2014, 11:42 AM
Post: #34
RE: Quiz: calculating a definite integral
(01-03-2014 07:03 AM)Thomas Klemm Wrote:  
(01-03-2014 06:08 AM)walter b Wrote:  there can be only one correct.
Use integration by parts to get: \(\int_{0}^{1}x^n\log(x)^p dx=-\frac{p}{n+1}\int_{0}^{1}x^n\log(x)^{p-1}dx\)
Iterate starting with \(p=n\) until you end up with: \(\int_{0}^{1}x^n dx=\frac{1}{n+1}\).
You can figure out by yourself which is correct.

Sorry, I made a mistake while copying from my paper: too late in the night and with a headhache ;-)The right expression is with n+1, it is now corrected.
That said 0⁰ is not a problem, it is quite easy to show that it is 1.

From a strict mathematical standpoint, I'm a bit puzzled as the Taylor expansion used seems a bit unjustified:
-the expansion of e^u(x) is not a sum of power of u(x) but a sum of successive derivatives of e^u(x) which doesn't seems to be the same.
-the first derivative of the function is not defined at 0, then the expansion can't converge, one could have used and expansion around 1 to go that way.

I do agree that the sum gives the right result but I don't grasp a rigorous proof.

If somebody has some light....

Regards
Find all posts by this user
Quote this message in a reply
Post Reply 


Messages In This Thread
RE: Quiz: calculating a definite integral - Bunuel66 - 01-03-2014 11:42 AM



User(s) browsing this thread: 1 Guest(s)