- #1

CAF123

Gold Member

- 2,888

- 88

## Main Question or Discussion Point

I want to evaluate $$\int_{a+b-c}^s\,\text{d}x\, \frac{(-x+ab/c)^{\epsilon}}{(x+c-a-b)^{\epsilon+1} (a-x)},$$ where ##a,b,c,\epsilon## are numbers, and to be treated as constants in the integration. I put this into mathematica and an hour later it is still attempting to evaluate it so I aborted the calculation. I was just wondering if someone could put the following code into Mathematica and see if it is also takes a long time on their machine?

Integrate[(-x + a*b/c)^e/(x + c - a - b)^(e + 1)/(a - x), {x,

a + b - c, s}, Assumptions -> e < 0 && b < 0 && a > 0]

I tried to progress with the intergral analytically:

Partial fractions gave a rewriting of the form $$\frac{1}{c-b}\int_{a+b-c}^{s} \text{d}x\, \frac{(-x+ab/c)^{\epsilon}}{(x+c-a-b)^{\epsilon}} \left(\frac{1}{x+c-a-b} + \frac{1}{a-x}\right)$$ $$ = \frac{1}{c-b}\left(\int_{a+b-c}^{s} \text{d}x\, \frac{(-x+ab/c)^{\epsilon}}{(x+c-a-b)^{\epsilon+1}} + \int_{a+b-c}^{s} \text{d}x\, \frac{(-x+ab/c)^{\epsilon}}{(x+c-a-b)^{\epsilon}} \frac{1}{a-x}\right)$$ but this probably did not help.

Thanks!

Integrate[(-x + a*b/c)^e/(x + c - a - b)^(e + 1)/(a - x), {x,

a + b - c, s}, Assumptions -> e < 0 && b < 0 && a > 0]

I tried to progress with the intergral analytically:

Partial fractions gave a rewriting of the form $$\frac{1}{c-b}\int_{a+b-c}^{s} \text{d}x\, \frac{(-x+ab/c)^{\epsilon}}{(x+c-a-b)^{\epsilon}} \left(\frac{1}{x+c-a-b} + \frac{1}{a-x}\right)$$ $$ = \frac{1}{c-b}\left(\int_{a+b-c}^{s} \text{d}x\, \frac{(-x+ab/c)^{\epsilon}}{(x+c-a-b)^{\epsilon+1}} + \int_{a+b-c}^{s} \text{d}x\, \frac{(-x+ab/c)^{\epsilon}}{(x+c-a-b)^{\epsilon}} \frac{1}{a-x}\right)$$ but this probably did not help.

Thanks!

Last edited by a moderator: