How to show the following equality? ∑n=0∞1a2+n2=1+aπcoth⁡aπ2a2

This problem is a direct application of Fourier transform and Poisson summation formula. Recalling the definition of Fourier transform and the Poisson summation formula respectively
${\displaystyle F\left(w\right)=\frac{1}{\sqrt{2\pi }}{\int }_{-\mathrm{\infty }}^{\mathrm{\infty }}f\left(x\right)e^{-xw}dx}$
${\displaystyle \sum _{-\mathrm{\infty }}^{\mathrm{\infty }}f\left(n\right)=\sqrt{2\pi }\sum _{-\mathrm{\infty }}^{\mathrm{\infty }}F\left(2n\pi \right)}$
where F is the Fourier transform of f. Advancing with our problem, first, we compute the Fourier transform of ${\displaystyle f\left(x\right)=\frac{1}{x^2+a^2}}$ wihich is equal to
${\displaystyle F\left(w\right)=\sqrt{\frac{\pi }{2}}\frac{1}{a}{e}^{-a\left|w\right|}}$
Applying Poisson formula, we have
${\displaystyle {\int }_{n=0}^{\mathrm{\infty }}\frac{1}{n^2+a^2}=\frac{\pi }{a}\sum _{n=0}^{\mathrm{\infty }}{e}^{-2an\pi }=\frac{\pi }{a}\sum _{n=0}^{\mathrm{\infty }}{r}^n}$
${\displaystyle \Rightarrow \sum _{n=0}^{\mathrm{\infty }}\frac{1}{n^2+a^2}=\frac{\pi }{a}\frac{1}{1-e^{-2a\pi }}}$
Now, I leave it to you to manipulate the above expression to reach the form
${\displaystyle \sum _{n=0}^{\mathrm{\infty }}\frac{1}{a^2+n^2}=\frac{1+a\pi \coth a\pi }{2a^2}}$
You can use the identity
${\displaystyle \coth x=\frac{\text{cosh}x}{\text{sinh}x}=\frac{e^x+e^{-x}}{e^x-e^{-x}}=\frac{e^{2x}+1}{e_1^{2x}}}$

It is well known that
${\displaystyle \sum _{n=-\mathrm{\infty }}^{\mathrm{\infty }}f\left(n\right)=-\sum _{j=1}^{k}Res\pi \cot \left(\pi z\right)f\left(z\right)}$
Assume ${\displaystyle a\ne 0}$
To find the residues of ${\displaystyle g\left(z\right)=\pi \cot \left(\pi z\right)\frac{1}{a^2+n^2}}$, we see
${\displaystyle \frac{1}{a^2+n^2}=\frac{1}{(n+ia)(n-ia)}}$
so g has poles at ${\displaystyle z_1=ia}$ and ${\displaystyle z_2=-ia}$. Their respective residues, ${\displaystyle b_1}$ and ${\displaystyle b_2}$ can be found
${\displaystyle b_1=Resg\left(z\right)=\underset{z\to ia}{lim}\pi \cot \left(\pi z\right)\frac{z-ia}{\}}\left\{(z+ia)(z-ia)\right\}=\pi \cot \left(\pi ia\right)\frac{1}{2ia}}$
And finally:
${\displaystyle \sum _{k=-\mathrm{\infty }}\frac{1}{a^2+k^2}=-(b_1+b_2)=\frac{\pi \coth \left(\pi a\right)}{a}}$
To change the starting number from ${\displaystyle -\mathrm{\infty }}$ to 0, we divide the series, as it is symmetrical (i.e. ${\displaystyle g\left(n\right)=g(-n)}$ ):
${\displaystyle \sum _{k=-\mathrm{\infty }}^{\mathrm{\infty }}\frac{1}{a^2+k^2}=\frac{\pi \coth \left(\pi a\right)}{a}=\sum _{k=-\mathrm{\infty }}^{-1}\frac{1}{a^2+k^2}+\frac{1}{a^2}+\sum _{k=1}^{\mathrm{\infty }}\frac{1}{a^2+k^2}}$

Thus
${\displaystyle \sum _{k=0}^{\mathrm{\infty }}\frac{1}{a^2+k^2}=\frac{\pi \coth \left(\pi a\right)}{2a}+\frac{1}{2a^2}=\frac{\pi \coth \left(\pi a\right)+1}{2a^2}}$