# Thread: Integral over Logarithm with Polynomial in the Argument

1. ## Integral over Logarithm with Polynomial in the Argument

Hello,
I would like to solve the following integral
$\displaystyle A(p^2 , m_R^2) = - \int dx \; ln(1 - x(1-x) \frac{p^2}{m_R^2})$

For $\displaystyle p^2 < (2 m_R)^2$ and for $\displaystyle p^2 > (2 m_R)^2$

I tried some substitutions without success.

Could anybody help me solving this?

Regards,
Mr. Fogg

2. Originally Posted by Kai.Grohs
Hello,
I would like to solve the following integral
$\displaystyle A(p^2 , m_R^2) = - \int dx \; ln(1 - x(1-x) \frac{p^2}{m_R^2})$

For $\displaystyle p^2 < (2 m_R)^2$ and for $\displaystyle p^2 > (2 m_R)^2$

I tried some substitutions without success.

Could anybody help me solving this?

Regards,
Mr. Fogg
Hi

Your problem is the same as solving this integral

$\displaystyle I = \int ln(ax^2+bx+c) \; dx$

In your problem the discriminant $\displaystyle b^2-4ac$ is negative if $\displaystyle p^2 < (2 m_R)^2$ and positive if $\displaystyle p^2 > (2 m_R)^2$

This integral can be solved by parts and then by rearranging a little bit into

$\displaystyle I = \left(x+\frac{b}{2a}\right) ln(ax^2+bx+c) - 2x - \int \frac{b^2-4ac}{2a(ax^2+bx+c)} \; dx$

The last integral depends by the sign of the discriminant

If negative then you can chose the proper change of variables to get (1+u²) at the denominator, and then integrate into arctan

If positive then you can factor the denominator into 2a²(x-x0)(x-x1) and go further in the decomposition of the last integral

3. Hello,
thanks for your help, but I don't see an integration by parts in this
$\displaystyle - \int_0^1 dx (1-x \frac{p^2}{m_R^2} + x^2 \frac{p^2}{m_R^2})$

I tried this
$\displaystyle - \int_0^1 dx \; x' (1-x \frac{p^2}{m_R^2} + x^2 \frac{p^2}{m_R^2})$
but it didn't help me. Maybe I made a mistake.

Could you help me a little more precisely?

Regards,
Kai

4. If you don't mind I will keep this form

$\displaystyle I = \int ln(ax^2+bx+c) \; dx$

and you will set $\displaystyle a = \frac{p^2}{m_R^2}$ etc ...

$\displaystyle I = \int ln(ax^2+bx+c) \; dx$

By parts with u'(x) = 1 and v(x) = ln(ax^2+bx+c)

$\displaystyle I = x\:ln(ax^2+bx+c) - \int\frac{x\2ax+b)}{ax^2+bx+c} \; dx$

$\displaystyle I = x\:ln(ax^2+bx+c) - \int\frac{2ax^2+2bx+2c-bx-2c}{ax^2+bx+c} \; dx$

$\displaystyle I = x\:ln(ax^2+bx+c) - \int \:2 \; dx + \int\frac{bx+2c}{ax^2+bx+c} \; dx$

$\displaystyle I = x\:ln(ax^2+bx+c) - 2\:x + \int\frac{bx+2c}{ax^2+bx+c} \; dx$

And

$\displaystyle \int\frac{bx+2c}{ax^2+bx+c} \; dx = \int\frac{\frac{b}{2a}(2ax+b)-\frac{b^2}{2a}+2c}{ax^2+bx+c} \; dx$

$\displaystyle \int\frac{bx+2c}{ax^2+bx+c} \; dx = \frac{b}{2a}\:ln(ax^2+bx+c) -\int \frac{\frac{b^2}{2a}-2c}{ax^2+bx+c} \; dx$

5. Originally Posted by running-gag
The last integral depends by the sign of the discriminant

If negative then you can chose the proper change of variables to get (1+u²) at the denominator, and then integrate into arctan

If positive then you can factor the denominator into 2a²(x-x0)(x-x1) and go further in the decomposition of the last integral
Could you tell me the proper change of variables. I don't see it. I am just a physicist, who has not this magic eye for finding the proper change

Regards,
Mr. Fogg

6. Let $\displaystyle \Delta = b^2-4ac$ be the discriminant
If $\displaystyle \Delta < 0$ then let $\displaystyle u = \frac{2a}{\sqrt{-\Delta}}\:\left(x+\frac{b}{2a}\right) = \frac{2a}{\sqrt{-\Delta}}\:x+\frac{b}{\sqrt{-\Delta}}$

Then

$\displaystyle - \int \frac{b^2-4ac}{2a(ax^2+bx+c)} \; dx = - \int \frac{b^2-4ac}{2a^2\left(x^2+\frac ba\:x+\frac ca\right)} \; dx$

$\displaystyle - \int \frac{b^2-4ac}{2a(ax^2+bx+c)} \; dx = - \int \frac{\Delta}{2a^2 \left[\left(x+\frac{b}{2a}\right)^2+\frac ca-\frac{b^2}{4a^2}\right]} \; dx$

$\displaystyle - \int \frac{b^2-4ac}{2a(ax^2+bx+c)} \; dx = - \int \frac{\Delta}{2a^2\left[\left(x+\frac{b}{2a}\right)^2-\frac{\Delta}{4a^2}\right]} \; dx$

$\displaystyle - \int \frac{b^2-4ac}{2a(ax^2+bx+c)} \; dx = - \int \frac{\Delta}{2a^2\left[-\frac{\Delta}{4a^2}\:u^2-\frac{\Delta}{4a^2}\right]} \; \frac{\sqrt{-\Delta}}{2a}\:du$

$\displaystyle - \int \frac{b^2-4ac}{2a(ax^2+bx+c)} \; dx = \frac{\sqrt{-\Delta}}{a}\:\int \frac{du}{1+u^2}$

$\displaystyle - \int \frac{b^2-4ac}{2a(ax^2+bx+c)} \; dx = \frac{\sqrt{-\Delta}}{a}\:Arctan(u)$

$\displaystyle - \int \frac{b^2-4ac}{2a(ax^2+bx+c)} \; dx = \frac{\sqrt{-\Delta}}{a}\:Arctan \left(\frac{2a}{\sqrt{-\Delta}}\:x+\frac{b}{\sqrt{-\Delta}}\right)$

7. Thanks for your help.

I went through your calculation, but didn't get the correct solution, which is

Do you have any idea, how to get this solution?

Ragards,
Mr. Fogg

8. So far I did not check the second case

But for the first case $\displaystyle p^2 < (2m_R^2)$ I applied the formula I gave you in my previous posts and I could find the good result

$\displaystyle a = \frac{p^2}{m_R^2}$

$\displaystyle b = -\frac{p^2}{m_R^2}$

$\displaystyle c = 1$

$\displaystyle \Delta = \left(\frac{p^2}{m_R^2}\right)^2-4\:\frac{p^2}{m_R^2} = \frac{p^2}{m_R^2}\:\left(\frac{p^2}{m_R^2}-4\right)$

In your problem the discriminant $\displaystyle b^2-4ac$ is negative if $\displaystyle p^2 < (2 m_R)^2$ and positive if $\displaystyle p^2 > (2 m_R)^2$

$\displaystyle I = \left(x+\frac{b}{2a}\right) ln(ax^2+bx+c) - 2x - \int \frac{b^2-4ac}{2a(ax^2+bx+c)} \; dx$

And

$\displaystyle - \int \frac{b^2-4ac}{2a(ax^2+bx+c)} \; dx = \frac{\sqrt{-\Delta}}{a}\:Arctan \left(\frac{2a}{\sqrt{-\Delta}}\:x+\frac{b}{\sqrt{-\Delta}}\right)$

Since you are integrating between 0 and 1 you do not have to care about the first term $\displaystyle \left(x+\frac{b}{2a}\right) ln(ax^2+bx+c)$ because for 1, ln(ax²+bx+c) gives ln(a+b+c)=ln(1)=0 ; and for 0, ln(ax²+bx+c) gives ln(c)=ln(1)=0

The second term (-2x) gives -2 for 1 and 0 for 0, therefore -2

For the third term $\displaystyle \frac{\sqrt{-\Delta}}{a}\:Arctan \left(\frac{2a}{\sqrt{-\Delta}}\:x+\frac{b}{\sqrt{-\Delta}}\right)$

For 1, $\displaystyle Arctan \left(\frac{2a}{\sqrt{-\Delta}}\:x+\frac{b}{\sqrt{-\Delta}}\right) = Arctan \left(\frac{2a+b}{\sqrt{-\Delta}}\right) = Arctan \left(\frac{a}{\sqrt{-\Delta}}\right)$

For 0, $\displaystyle Arctan \left(\frac{2a}{\sqrt{-\Delta}}\:x+\frac{b}{\sqrt{-\Delta}}\right) = Arctan \left(\frac{b}{\sqrt{-\Delta}}\right) = Arctan \left(\frac{-a}{\sqrt{-\Delta}}\right)$

But Arctan(-x)=-Arctan(x) therefore for 0 it gives $\displaystyle -Arctan \left(\frac{a}{\sqrt{-\Delta}}\right)$

Therefore the third term gives $\displaystyle 2\:\frac{\sqrt{-\Delta}}{a}\:Arctan \left(\frac{a}{\sqrt{-\Delta}}\right)$

$\displaystyle \frac{\sqrt{-\Delta}}{a} = \frac{\sqrt{4\:\frac{p^2}{m_R^2}-\left(\frac{p^2}{m_R^2}\right)^2}}{\frac{p^2}{m_R^ 2}}= \frac{m_R^2}{p^2}\:\sqrt{4\:\frac{p^2}{m_R^2}-\frac{p^4}{m_R^4}}$

$\displaystyle \frac{\sqrt{-\Delta}}{a} = \sqrt{\frac{m_R^4}{p^4}\:\left(4\:\frac{p^2}{m_R^2 }-\frac{p^4}{m_R^4}\right)} = \sqrt{4\:\frac{m_R^2}{p^2}-1} = \sqrt{\frac{(2m_R)^2}{p^2}-1}$

At the end the third term gives

$\displaystyle 2\:\sqrt{\frac{(2m_R)^2}{p^2}-1}\:\:Arctan \left(\frac{1}{\sqrt{\frac{(2m_R)^2}{p^2}-1}}\right) = 2\:\left(\frac{(2m_R)^2}{p^2}-1\right)^{\frac12}\:\:tan^{-1} \left(\frac{(2m_R)^2}{p^2}-1\right)^{-\frac12}$