User Portlet
| Discussions |
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| OK, WorkingPrecision works, but not AccuracyGoal. What's the difference i.e. why would AccuracyGoal refuse to produce more significant digits, is it because WorkingPrecision is some sort of brute force? |
| Thank you David. |
| Yes, here is a Notebook with 3 Integrals, two of them with just a difference in symbol ("q1" instead of "a"). Also (reply to another post) the Integrals are not all necessarily divergent, all variables are Real and greater than 0. |
| Hennrik, On a completely separate note, how do you typset your post to include such nice math symbols? |
| Yes Bill, that's it. I have no idea why my browser can't render the Mathematica display (firefox and chrome). |
| Thank you Marco! [Thanks everyone who replied] This is exactly what I needed: Question: Is the command "Manipulate" enough/essential to check where the graph exists i.e. what the constraints on the three variables are? |
| Assuming[{p, m, \[CapitalLambda], k} \[Element] Reals, Integrate[k/(2 (k^2 + m^2) (-k + p)), {p, 0, \[CapitalLambda]}]] Mathematica's answer is ConditionalExpression[(k Log[1 - \[CapitalLambda]/k])/(2 (k^2 + m^2)), \[CapitalLambda] == 0 ||... |
| The integrals (especially the first one) is pretty straightforward if done by partial fractions -- indicated in my earlier mail. p^2/(p^2 - k^2) = (p^2 + k^2 - k^2)/((p + k) (p - k)) = 1 + k^2/((p + k) (p - k)) For certain cases where Em |
| Appreciate tips on how write a Mathematica script that would 1. Wait for user input, that contains a long string which includes blank spaces, i.e. no commas etc. 2. Do an integral over a predefined variable using input from Step 1. 3. Use... |
| Hello, There is no problem solving this integral: Integrate[DiracDelta[k1 - pm] DiracDelta[-k1 + q1] w[pm, m],{pm,-Infinity,Infinity}]which gives [mcode]ConditionalExpression[DiracDelta[-k1 + q1] w[k1, m], k1 \[Element]... |