Timeline for Trouble evaluating the following integral
Current License: CC BY-SA 4.0
5 events
| when toggle format | what | by | license | comment | |
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| Nov 14, 2023 at 6:48 | comment | added | Bob Hanlon |
Since Assuming[m > 0 && Q > 0, Log[m^2 - Q^2 x (1 - x)] == Log[m^2*(1 - (Q/m)^2 x (1 - x))] // Simplify] evaluates to True; just define rule = Log[m^2*(1 - (Q/m)^2 x (1 - x))] :> Log[1 - (Q/m)^2 x (1 - x)] + Log[m^2]; and use expr /. rule as required.
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| Nov 14, 2023 at 4:40 | comment | added | MathZilla | Also, when I put this on my bigger computer, it does the entire integral in about 1 minute. The condition Mathematica puts on the integral is that $2m>Q$ and $Q\leq Re[\sqrt{-4m^2+Q^2}]$. | |
| Nov 14, 2023 at 4:31 | comment | added | MathZilla | Is there a way to expand the $\log(m^2-Q^2x(1-x))$ within Mathematica such that I get the following output, $\log(1-(Q/m)^2x(1-x)) + log(m^2)$ or for those of with the inverse inside the log, write it as $\log(1/(m^2 - Q^2x(1-x)) = -\log(m^2 - Q^2x(1-x))$? Or would this not even help the situation? | |
| Nov 13, 2023 at 23:01 | vote | accept | MathZilla | ||
| Nov 13, 2023 at 22:40 | history | answered | Bob Hanlon | CC BY-SA 4.0 |