0
|
9048 Views
|
5 Replies
|
3 Total Likes
View groups...
Share
GROUPS:

# [?] Solve this indefinite integral?

Posted 6 years ago
 I evaluate $$\int \frac{1}{y (y-1)} \, dy$$ which I expect to give $$\log (y-1)-\log (y)$$ and I get $$\log (1-y)-\log (y)$$ Is this a bug?
5 Replies
Sort By:
Posted 6 years ago
 It's still good to have A. P. Prudnikov, Yu. A. Brychkov, O. I. Marichev (who works since years with Wolfram): Integrals and Series (Moscow 1981, russian) on the book shelf, and there in ยง 1.2.5 no. 9, p. 31: Integrate[1/(x (x + a)), x] = Log[Abs[x/(x+a)]]/a it's the same thing as discussed in Doing an integral ... mentioned above. To make a long story short: use the Prudnikov et. al. solution as long as $x$ is real.Two of the integral table workers (Brychkow and Marichev) wrote an interesting blog article New Derivatives of the Bessel Functions Have Been Discovered with the Help of the Wolfram Language! Enjoy!
Posted 6 years ago
 Thank you very much. That is a very interesting and useful post. (Doing an integral ...)In effect, Mathematica leaves out the Abs[]
Posted 6 years ago
 Thank you very much. Yes, it never occurred to me that D[Log[1 - y], y] could be the same as D[Log[y - 1], y]Sorry for not stopping to think.I guess the reason Mathematica chooses the first anti-derivative is because of its preference for putting terms in $y^0$ first.If I had been dealing with y's less than 1, I would have been happy with my result...getting negative numbers inside the logarithm was a bit of a shock (not that I was actually evaluating them, but I could see that for the part of the parameter space I was dealing with it was a problem).
Posted 6 years ago
 Is this a bug? No this is a correct anti-derivative because of In[2]:= Together[D[Log[1 \[Minus] y] \[Minus] Log[y], y]] Out[2]= 1/((-1 + y) y) Read about another correct anti-derivative here Doing an integral: From Mathematica Version 1.0 to 11.0