Evaluate the following integral from a physics textbook

[dRic2]

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 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

 

[SammyS]

Is it something as basic as $\displaystyle \int {\frac{1}{x}} dx = \log|x| + C$ ?

 

[dRic2]

Mhm I'm embarrassed that I didn't thought of that, thanks!

BTW integrating $\log |x+a|$ seems a pain... I'm going to see what I can do

 

[SammyS]

Mhm I'm embarrassed that I didn't thought of that, thanks!

BTW integrating $\log |x+a|$ seems a pain... I'm going to see what I can do

First, try integrating $\log|x |$, then go for integrating $\log |x+a|$ .

It could be done by trial & error.

Better: It can be done by Integration by Parts. \ _ |
You might say, What parts are possible in $\int \log|x |\ dx$ ?
Seems like the only possibility is: Let $u= \log|x |$ and $dv = dx$

Then $du= \dfrac{1}{x} dx$ and $v = x$ .​

Of course, you can look it up in table of integration formulas.

 

[dRic2]

Sorry I was traveling yesterday so I couldn't try. I'll be back home on Monday.

BTW I know how to integrate log(x)... It is the absolute value that annoys me... It might be actually easier, but I'm always suspicious whenever I encounter an absolute value because it always screw up the domain of the function :D (this times seems to make it easier though)

Ps: the funniest way to integrate log(x) is to integrate by parts $\int (1)* \log(x) = x \log(x) - \int x \frac 1 x$

 

[dRic2]

For example. Take the above example. It gets harder since the derivative of $\log |x|$ is no longer just $\frac 1 x$ but instead $sgn(x) \frac 1 x$. So the integral of $\log |x| = x \log|x| - sgn(x)x$ which depends of the limit of integration since sgn(x) could be 1 or -1... It is a pain... at least to me :D

 

[vela]

For example. Take the above example. It gets harder since the derivative of $\log |x|$ is no longer just $\frac 1 x$ but instead $sgn(x) \frac 1 x$.

This isn't correct. $\log \lvert x \rvert$ monotonically decreases on the interval $(-\infty,0)$, so its derivative should be negative, which is exactly what $1/x$ gives you.

 

[dRic2]

Sorry I messed up. It is $sgn(x) \frac 1 {|x|} = \frac 1 x$. So I guess I was worring for nothing...

 

[dRic2]

So here I thought to use

∫klog(k+a)=k∫log(k+a)−∫log(k+a)=(k−1)∫log(k+a)=...∫klog⁡(k+a)=k∫log⁡(k+a)−∫log⁡(k+a)=(k−1)∫log⁡(k+a)=...​

\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.

(Sorry about the latext)

This is nonsense. I apologize. I've made a mistake forgetting to integrate log(k+a) in the second part. Just wanted to be clear for other people interested in the post. A different integration by parts is needed:
$$\int x log(x+a) = x^2 log(x+a) - \int \frac {x^2} {x+a}$$
The one proceeds by splitting x^2/(x+a) into x - (ax)/(x+a) and so on...