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The sum is only convergent for $p>1/2$, so for $p=1/2$ you could regularize it by adding a small positive increment: $p=1/2+\epsilon$, $\epsilon>0$.

For $a\gg 1$ you can then approximate the sum by an integral, to arrive at

$$I_{a,p}=\int_0^\infty \frac{1}{(a+x^2)^p}dx=\frac{\sqrt{\pi a } \Gamma (p-1/2)}{2 a^p\Gamma (p)},\;\;\text{for}\;\;p>1/2.$$

This compares quite well with the sum

$$S_{a,p}=\sum_{n=0}^\infty\frac{1}{(a+n^2)^p}$$

as you can see from the plot where both $I_{a,p}$ (blue) and $S_{a,p}$ (orange) are plotted as a function of $a$ for $p=0.51$.

This is for $p=1/2+\epsilon$. For the larger values of $p$ you mention no regularization is needed.

The sum is only convergent for $p>1/2$, so for $p=1/2$ you could regularize it by adding a small positive increment: $p=1/2+\epsilon$, $\epsilon>0$.

For $a\gg 1$ you can then approximate the sum by an integral, to arrive at

$$I_{a,p}=\int_0^\infty \frac{1}{(a+x^2)^p}dx=\frac{\sqrt{\pi a } \Gamma (p-1/2)}{2 a^p\Gamma (p)},\;\;\text{for}\;\;p>1/2.$$

This compares quite well with the sum

$$S_{a,p}=\sum_{n=0}^\infty\frac{1}{(a+n^2)^p}$$

as you can see from the plot where both $I_{a,p}$ (blue) and $S_{a,p}$ (orange) are plotted as a function of $a$ for $p=0.51$.

This is for $p=1/2+\epsilon$. For the larger values of $p$ you mention no regularization is needed.

The sum is only convergent for $p>1/2$, so for $p=1/2$ you could regularize it by adding a small positive increment: $p=1/2+\epsilon$, $\epsilon>0$.

For $a\gg 1$ you can then approximate the sum by an integral, to arrive at

$$I_{a,p}=\int_0^\infty \frac{1}{(a+x^2)^p}dx=\frac{\sqrt{\pi a } \Gamma (p-1/2)}{2 a^p\Gamma (p)},\;\;\text{for}\;\;p>1/2.$$

This compares quite well with the sum

$$S_{a,p}=\sum_{n=0}^\infty\frac{1}{(a+n^2)^p}$$

as you can see from the plot where both $I_{a,p}$ (blue) and $S_{a,p}$ (orange) are plotted as a function of $a$ for $p=0.51$.

This is for $p=1/2+\epsilon$. For the larger values of $p$ you mention no regularization is needed and the sum can be evaluated directly, for example

$$S_{a,1}=\frac{\pi^2}{2\alpha^2}(1+\alpha\coth\alpha),\;\;\alpha\equiv\pi\sqrt a,$$ $$S_{a,3/2}=\frac{\pi^6}{32 \alpha^{6}}\left(2\alpha^3\coth \alpha\sinh^{-2}\alpha+3 \alpha\coth \alpha+3 \alpha^2 \,\sinh^{-2}\alpha +8\right).$$

The sum is only convergent for $p>1/2$, so for $p=1/2$ you could regularize it by adding a small positive increment: $p=1/2+\epsilon$, $\epsilon>0$.

For $a\gg 1$ you can then approximate the sum by an integral, to arrive at

$$I_{a,p}=\int_0^\infty \frac{1}{(a+x^2)^p}dx=\frac{\sqrt{\pi a } \Gamma (p-1/2)}{2 a^p\Gamma (p)},\;\;\text{for}\;\;p>1/2.$$

This compares quite well with the sum

$$S_{a,p}=\sum_{n=0}^\infty\frac{1}{(a+n^2)^p}$$

as you can see from the plot where both $I_{a,p}$ (blue) and $S_{a,p}$ (orange) are plotted as a function of $a$ for $p=0.51$.

This is for $p=1/2+\epsilon$. For the larger values of $p$ you mention no regularization is needed.

The sum is only convergent for $p>1/2$, so for $p=1/2$ you could regularize it by adding a small positive increment: $p=1/2+\epsilon$, $\epsilon>0$.

For $a\gg 1$ you can then approximate the sum by an integral, to arrive at

$$I_{a,p}=\int_0^\infty \frac{1}{(a+x^2)^p}dx=\frac{\sqrt{\pi a } \Gamma (p-1/2)}{2 a^p\Gamma (p)},\;\;\text{for}\;\;p>1/2.$$

This compares quite well with the sum

$$S_{a,p}=\sum_{n=0}^\infty\frac{1}{(a+n^2)^p}$$

as you can see from the plot where both $I_{a,p}$ (blue) and $S_{a,p}$ (orange) are plotted as a function of $a$ for $p=0.51$.

This is for $p=1/2+\epsilon$. For the larger values of $p$ you mention no regularization is needed and the sum can be evaluated directly, for example

$$S_{a,1}=\frac{\pi^2}{2\alpha^2}(1+\alpha\coth\alpha),\;\;\alpha\equiv\pi\sqrt a,$$ $$S_{a,3/2}=\frac{\pi^6}{32 \alpha^{6}}\left[2\alpha^3\coth \alpha\sinh^{-2}\alpha+3 \alpha\coth \alpha+3 \alpha \,\sinh^{-2}\alpha +8\right].$$

The sum is only convergent for $p>1/2$, so for $p=1/2$ you could regularize it by adding a small positive increment: $p=1/2+\epsilon$, $\epsilon>0$.

For $a\gg 1$ you can then approximate the sum by an integral, to arrive at

$$I_{a,p}=\int_0^\infty \frac{1}{(a+x^2)^p}dx=\frac{\sqrt{\pi a } \Gamma (p-1/2)}{2 a^p\Gamma (p)},\;\;\text{for}\;\;p>1/2.$$

This compares quite well with the sum

$$S_{a,p}=\sum_{n=0}^\infty\frac{1}{(a+n^2)^p}$$

as you can see from the plot where both $I_{a,p}$ (blue) and $S_{a,p}$ (orange) are plotted as a function of $a$ for $p=0.51$.

This is for $p=1/2+\epsilon$. For the larger values of $p$ you mention no regularization is needed and the sum can be evaluated directly, for example

$$S_{a,1}=\frac{\pi^2}{2\alpha^2}(1+\alpha\coth\alpha),\;\;\alpha\equiv\pi\sqrt a,$$ $$S_{a,3/2}=\frac{\pi^6}{32 \alpha^{6}}\left(2\alpha^3\coth \alpha\sinh^{-2}\alpha+3 \alpha\coth \alpha+3 \alpha^2 \,\sinh^{-2}\alpha +8\right).$$