# Possible integration by parts?

1. Dec 7, 2015

### CAF123

1. The problem statement, all variables and given/known data
Integrate $$\int_0^1 dw \frac{w^{\epsilon+1} \ln((r+1-w)/r)}{1+r(1+w)}$$ for $\epsilon$ not necessarily an integer but positive and r is negative (<-1). The argument of the log is positive.

2. Relevant equations
Integration by parts

3. The attempt at a solution

I can only think of integration by parts, differentiating the log to bring back the rational but I wouldn't be able to integrate the remaining piece of the integrand. I suspect the result will be a hypergeometric function but I can't seem to get there. I had a similar integral with $\ln (w)$ instead and my prof suggested I write this as $$\ln w = \frac{d}{d\beta} w^{\beta}|_{\beta=0} = \frac{d}{d\beta} e^{\beta \ln w}|_{\beta=0}$$ but I am not sure why this helps.

Thanks for any help!

2. Dec 7, 2015

### Staff: Mentor

Did you test it?
It gets rid of the logarithm. You have to integrate with respect to β later, but usually this is an easy integral.

3. Dec 8, 2015

### CAF123

Hi mfb,
The integral was $$\int_0^1 dw \frac{w^{\epsilon+1}}{(1+w)(1+r(1+w))} \ln w = \int_0^1 dw \frac{w^{\epsilon+1}}{(1+w)(1+r(1+w))} \frac{d}{d \beta} w^{\beta}|_{\beta=0}$$ Since there is only one instance of a beta dependence I can write $$\frac{d}{d \beta}|_{\beta=0} \int_0^1 dw \frac{w^{\epsilon+1+\beta}}{(1+w)(1+r(1+w))}$$ This is where I didn't see how the rewriting of the log helped. The integrand nearly has a hypergeometric structure but, even if I identified it as a hypergeometric, I still have a derivative wrt beta outside. I could proceed with integration by parts more easily now, but the integral I get from doing so is again just as complicated so I would hit a dead end again. Any thoughts?

4. Dec 8, 2015

### Staff: Mentor

The first integral in the first post was different than the one in your second post. Which one is right?

5. Dec 9, 2015

### CAF123

There are two integrals I am trying to solve: $$\int_0^1 dw \frac{w^{\epsilon+1}}{1+r(1+w)} \ln\left(\frac{r+1-w}{r}\right),$$ (as in the OP) and $$\int_0^1 dw \frac{w^{\epsilon+1}}{(w+1)(1+r(w+1))} \ln w$$ It was for this latter integral that my prof suggested I use the rewriting of the log but, while I see that it leaves us with just rational factors, I don't see why it leads to a solution of the integral and if such a rewriting is of help for the integral I posted in the OP. Thanks!

6. Dec 9, 2015

### Staff: Mentor

Hmm...
Feynman parametrization? I don't know, it is just an idea that might work. The trick to substitute the log is similar.