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Akku14
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Do integration first and then take the limit k-> inf.

Use indefinite integration. Integrate only finds a solution for m==0.

g[x_, m_, k_] = 1/((2*(x - m))^(2*k) + 1)

mint[x_, k_] = Integrate[g[x, 0, k], x]

(*   x Hypergeometric2F1[1, 1/(2 k), 1 + 1/(2 k), -4^k x^(2 k)]   *)

Use a trick. Tell Limit that x^(2 k) is always positive.

mint2[x_, k_] = mint[x, k] /. x^(2 k) -> Abs[x]^(2 k)

Limit[mint2[1/2, k] - mint2[-1/2, k], k -> \[Infinity]]

(*   1   *)

Graphic shows the same.

Manipulate[Plot[mint[x, k], {x, -.5, .5}], {k, 1, 1000}]

Rubi (https://rulebasedintegration.org/) does the integral with arbitrary m.

rint[x_, m_, k_] = Int[g[x, m, k], x]

(*   (-m + x) Hypergeometric2F1[1, 1/(2 k), 
       1/2 (2 + 1/k), -4^k (-m + x)^(2 k)]   *)

rint2[x_, m_, k_] = 
    rint[x, m, k] /. (-m + x)^(2 k) -> Abs[(-m + x)]^(2 k)

Limit[rint2[m + 1/2, m, k] - rint2[m - 1/2, m, k], k -> \[Infinity]]

(*   1   *)
Akku14
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