Solving Tough Integral with Log - Need Help

[Hepth]

I posted this in the mathematical/computation software forum but maybe there's an algebraic trick I don't know to help me solve this.

I need to compute an integral of:
$$\sqrt{\text{E1}^2-m^2} Log\left( \frac{m^2-m_B E1-m_B\sqrt{E1^2-m^2}}{m^2-m_B E1+m_B \sqrt{E1^2-m^2}}\right)$$

over the region {E1,m,mB/2}

the limits of integration are well defined in the equation, though the function is asymptotic near the upper bound. But it still has a value there.

I can't seem to integrate this. Mathematica just spits the input back out. I know its possible because I can numerically integrate it for what I want and it gives a reasonable answer. Unfortunately I REALLY want an algebraic solution...

Does anyone have any ideas? An integral of a sqrt times a log of a function of the variable. I've tried some substitution methods, nothing seems to simplify.

Thanks for the help!


EDIT:I think I got something by letting U= Log[everything], it can be algebraic, but its sooo long... any other choices?

 

[Redbelly98]

I posted this in the mathematical/computation software forum but maybe there's an algebraic trick I don't know to help me solve this.

I need to compute an integral of:
$$\sqrt{\text{E1}^2-m^2} Log\left( \frac{m^2-m_B E1-m_B\sqrt{E1^2-m^2}}{m^2-m_B E1+m_B \sqrt{E1^2-m^2}}\right)$$

over the region {E1,m,mB/2}

What is the variable here? Why are there three integration limits {E1,m,mB/2} instead of two?

 

[Hepth]

variable is E1. From m to mB/2

I used mathematicas way of saying that sorry.

 

[Count Iblis]

EDIT:I think I got something by letting U= Log[everything], it can be algebraic, but its sooo long... any other choices?

Well, you can also use Mathematica do perform these steps and simplify the end result...

 

[Ben Niehoff]

It looks like you might be able to rewrite the logarithm in terms of inverse trigonometric (or hyperbolic) functions. Not sure if that helps you do the integral, though. Also, did you try giving Mathematica some Assumptions as to the ranges of the variables? Sometimes if it knows a variable is real and positive, it is able to simplify, because now it knows what branch every function needs to be evaluated on.