I get this as output:
$\frac{1}{6} \int_0^{\infty } 6 t \, dt$
Why it cannot cancel the coefficients?
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- 1$\begingroup$ Please, provide the code you have used that motivated this question. Latex is not so useful here. $\endgroup$CA Trevillian– CA Trevillian2021-03-21 10:19:04 +00:00Commented Mar 21, 2021 at 10:19
- $\begingroup$ Strictly speaking the coefficients should cancel only if the integral is well defined. $\endgroup$A.G.– A.G.2021-03-21 11:00:09 +00:00Commented Mar 21, 2021 at 11:00
- $\begingroup$ Divergence is not necessary here. You have the same behaviour with any integral Mathematica cannot evaluate. $\endgroup$mikado– mikado2021-03-21 11:14:43 +00:00Commented Mar 21, 2021 at 11:14
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1 Answer
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As much as I know, Mma does not simplify any integrals. However, one can help it. Try the following. Let us first introduce a rule:
intRule = Inactivate[Integrate[exprA___ exprB__ , {var_, lim__}], Integrate] /; FreeQ[{exprB}, var] :> exprB Inactivate[Integrate[exprA , {var, lim}], Integrate]; Here is your integral (inactivated):
expr1 = Inactivate[1/6*Integrate[6*t, {t, 0, \[Infinity]}], Integrate] (* 1/6 Inactive[Integrate][6 t, {t, 0, \[Infinity]}] *) Now
expr1 /. intRule (* Inactive[Integrate][t, {t, 0, \[Infinity]}] *) 
Have fun!
