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0

It's actually rather easy to specify the evenness or oddness of an integer in an assumption: Assuming[Mod[k, 2] == 0, Integrate[Sin[x]^k, {x, 0, 2 π}]] (* (2 Sqrt[π] Gamma[(1 + k)/2])/Gamma[1 + k/2] *) Assuming[Mod[k, 2] == 1, Integrate[Sin[x]^k, {x, 0, 2 π}]] (* 0 *)

1

Assuming[{n/2 ∈ Integers}, Integrate[Sin[x]^n, {x, 0, 2 π}]] // TraditionalForm Assuming[{n >= 0 && n/2 ∈ Integers}, Integrate[Sin[x]^n, {x, 0, 2 π}]] // TraditionalForm Assuming[{n >= 0 && n/2 ∈ Integers}, Integrate[Sin[x]^n, {x, 0, 2 π}] // FullSimplify] // TraditionalForm

0

I tried: Integrate[Sin[x]^(2 n), {x, 0, 2 Pi}, Assumptions -> {n >= 0 && n \[Element] Integers}] //TraditionalForm $$\frac{\sqrt{\pi } \left((-1)^{2 n}+1\right) \Gamma \left(n+\frac{1}{2}\right)}{\Gamma (n+1)}$$

7

{x,y} > 0 is not doing what you think it does. In contrast to Element which, as its documentation states, accepts arbitrary patterns and in which first argument, being a List of more than one elements, evaluates to Alternatives: $Assumptions = {x, y} ∈ Reals (* (x | y) ∈ Reals *) x ∈ Reals // Refine (* True *) y ∈ Reals // Refine (* True *) {x,y} > ... 3$Assumptions = And @@ Thread[Greater[{x, y, d, e}, 0]] && Element[{x, y, d, e}, Reals] x > 0 && y > 0 && d > 0 && e > 0 && (x | y | d | e) ∈ Reals ComplexExpand[Im[1/(y^2)^(9/2)]] 0

0

This has been answered before here Try Refine[Conjugate[f[x]], J \[Element] Reals]

0

FullSimplify@Assuming[n > 0, Integrate[ UnitBox[2 (x - y)] UnitBox[n y], {y, -\[Infinity], +\[Infinity]}]] \$\begin{cases} \frac{1}{2} & (n>0\land 4 n x+2>n\land 4 n x+n\leq 2)\lor (n=2\land x=0) \\ \frac{1}{n} & n>2\land 4 n x+n\geq 2\land 4 n x+2\leq n \\ \frac{1}{2}-x & n=2\land 0<x<\frac{1}{2} \\ x+\frac{1}{2} ...

1

Although, as the OP noted, Maximize[{Log[x] + Log[y], x + y == 10 && 0 < x && 0 < y}, {x, y}, Reals] returns unevaluated, Maximize[{Log[x] + Log[y], x + y == 10 && 0. < x && 0. < y}, {x, y}, Reals] (* {3.21888, {x -> 5., y -> 5.}} *) works. On the other hand, Maximize[{Log[x] + Log[y], x + y == 10 ...

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