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When I type(a^2)^s Mathematica does not give me $a^{2s}$ instead it gives ${(a^2)}^s$. Is there a way to make it print $a^{2s}$.

It made some real difference where I wanted to compute

Sum[((n \[Pi]/T)^2)^-s, {n, 1, Infinity}]

vs

Sum[((n \[Pi]/T))^(-2 s), {n, 1, Infinity}]

In the first case it gave me $\sum _{n=1}^{\infty } \pi ^{-2 s} \left(\frac{n^2}{T^2}\right)^{-s}$ but in the second case it gave me $\pi ^{-2 s} \left(\frac{1}{T}\right)^{-2 s} \zeta (2 s)$

I had to take the s-derivative of the result so I would like to get the result like the second one.

So, my question:

Is there any Mathematica function which helps me make Mathematica evaluate the first command and give back me the result like that of the second one?

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    $\begingroup$ Compare (a^2)^s /. {a -> -1, s -> 1/2} against a^(2 s) /. {a -> -1, s -> 1/2}. Also consider Simplify[(a^2)^s, Assumptions -> a > 0]. $\endgroup$
    – JimB
    Commented Feb 6, 2021 at 4:35
  • $\begingroup$ (a^2)^s // PowerExpand and Sum[((n \[Pi]/T)^2)^-s // PowerExpand, {n, 1, Infinity}] works. $\endgroup$
    – Bill Watts
    Commented Feb 6, 2021 at 6:39
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    $\begingroup$ To check if you are making some implied assumptions, use FindInstance to check for counterexamples, e.g., FindInstance[(a^2)^s != a^(2s), {a, s}, 5] and FindInstance[(a^2)^s != a^(2s), {a, s}, Reals, 5] $\endgroup$
    – Bob Hanlon
    Commented Feb 6, 2021 at 13:42

1 Answer 1

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FullSimplify[Sum[((n \[Pi]/T)^2)^-s, {n, 1, Infinity}],Assumptions->{T>0}]

Take your First input and assume $T>0$, that gives you the same result as the second one.

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