# Evaluate the following integral from a physics textbook

• dRic2
In summary: This is nonsense. I apologize. I've made a mistake forgetting to integrate log(k+a) in the second part.
dRic2
Gold Member
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

Delta2
Is it something as basic as ##\displaystyle \int {\frac{1}{x}} dx = \log|x| + C ## ?

Last edited:
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

dRic2 said:
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.

jim mcnamara
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##

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

dRic2 said:
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 said:
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.

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...

## 1. What is an integral in physics?

An integral in physics is a mathematical operation that is used to calculate the area under a curve on a graph. It is a fundamental concept in physics and is often used to solve problems related to motion, forces, and energy.

## 2. Why is it important to evaluate integrals in physics?

Evaluating integrals in physics is important because it allows us to calculate important physical quantities such as displacement, velocity, acceleration, work, and energy. These quantities are essential for understanding and predicting the behavior of objects in the physical world.

## 3. How do you evaluate an integral from a physics textbook?

To evaluate an integral from a physics textbook, you need to follow a set of steps, which may vary depending on the specific problem. Generally, you need to identify the limits of integration, determine the appropriate integration technique, and apply the fundamental theorem of calculus to solve the integral.

## 4. What are some common integration techniques used in physics?

Some common integration techniques used in physics include substitution, integration by parts, and partial fractions. These techniques allow us to solve a wide range of integrals that appear in physics problems.

## 5. Can integrals be used in other branches of science?

Yes, integrals are a fundamental tool in mathematics and are used in many other branches of science, including chemistry, biology, economics, and engineering. They allow us to model and analyze a wide range of phenomena in these fields.

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