- #1

dRic2

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- Homework Statement
- Hi, I would like to evaluate the following integral from a physics textbook:

$$-P\int d^3 k \theta (1 - k) \frac 2 { qk \cos \phi + \frac 1 2 q^2 }$$

(with P I mean the Cauchy principal value and ##\theta(x)## is the step function)

- Relevant Equations
- .

Using spherical coordinates I can write ##d^3 k = 2\pi k^2 \sin \phi dk d \phi## (where I've already preformed the integration along the azimuthal angle, yielding the factor ##2 \pi##). Btw I'm sorry for my unfortunate notation: usually ##\phi## is the azimuthal angle, but here it is the polar angle because I used ##\theta## for the step function...

Using the substitution ##\cos \phi = \eta## I can re-write the integral as:

$$- 4 \pi P\int_0^1 dk k^2 \int_{-1}^{1} d\eta \frac {qk} { qk \eta + \frac 1 2 q^2 } \frac {1} {qk}$$

(where I have multiplied by ##\frac {qk}{qk}## and used ##\theta(1-k)## to modify the limits of integration of the variable ##k##). Ok, so I do not really know much about the theory of Cauchy principal value, so I treated this as a normal integral and I got (integrating over ##\eta##):

$$- 4 \pi P\int_0^1 dk\log \left( \frac {qk + \frac 1 2 q^2 } {-qk + \frac 1 2 q^2 } \right) \frac {k} {q} = - 4 \pi P\int_0^1 dk\log \left({k + \frac 1 2 q }\right) \frac {k} {q} + 4 \pi P\int_0^1 dk\log \left( {-k + \frac 1 2 q } \right) \frac {k} {q}$$

(in the last step I simplified a ##q## because both ##k## and ##q## are

$$ \int k \log(k+a) = k \int \log(k+a) - \int \log(k+a) = (k-1)\int \log(k+a) = ...$$ and from now on it's mostly just algebra.

So my problem is that the solution should be this one (never mind all the missing constants in front of the integral):

$$\left[ -1 + \frac 1 q (1 - \frac 1 4 q^2) \log \left| \frac {2 - q} {2 + q} \right| \right]$$

And my question is: how am I supposed to get the absolute value in the logarithm ? It can't follow from just algebra... there is something from the theory I'm missing right?

Thanks

Ric

Using the substitution ##\cos \phi = \eta## I can re-write the integral as:

$$- 4 \pi P\int_0^1 dk k^2 \int_{-1}^{1} d\eta \frac {qk} { qk \eta + \frac 1 2 q^2 } \frac {1} {qk}$$

(where I have multiplied by ##\frac {qk}{qk}## and used ##\theta(1-k)## to modify the limits of integration of the variable ##k##). Ok, so I do not really know much about the theory of Cauchy principal value, so I treated this as a normal integral and I got (integrating over ##\eta##):

$$- 4 \pi P\int_0^1 dk\log \left( \frac {qk + \frac 1 2 q^2 } {-qk + \frac 1 2 q^2 } \right) \frac {k} {q} = - 4 \pi P\int_0^1 dk\log \left({k + \frac 1 2 q }\right) \frac {k} {q} + 4 \pi P\int_0^1 dk\log \left( {-k + \frac 1 2 q } \right) \frac {k} {q}$$

(in the last step I simplified a ##q## because both ##k## and ##q## are

*positively defined quantities*(sorry for not mentioning it earlier). So here I thought to use$$ \int k \log(k+a) = k \int \log(k+a) - \int \log(k+a) = (k-1)\int \log(k+a) = ...$$ and from now on it's mostly just algebra.

So my problem is that the solution should be this one (never mind all the missing constants in front of the integral):

$$\left[ -1 + \frac 1 q (1 - \frac 1 4 q^2) \log \left| \frac {2 - q} {2 + q} \right| \right]$$

And my question is: how am I supposed to get the absolute value in the logarithm ? It can't follow from just algebra... there is something from the theory I'm missing right?

Thanks

Ric