<s>[INST] Derivation of $1/(s^2-a^2)^{3/2}$ in Laurent Series So Given modified bessel equation (t/a)*I_1(at) using the general formula 
I have arrived at the answer.
$$\frac{1}{s^3}+\frac{3a^2}{2s^5}+\frac{15a^4}{8s^7}+ \cdots$$
And according to wolfram alpha its laurent series is equal to $\frac{1}{(s^2-a^2)^{3/2}}$ equal to my answer above
So the question is how my answer
$$\frac{1}{s^3}+\frac{3a^2}{2s^5}+\frac{15a^4}{8s^7} +
\cdots$$
equal to:
$$\frac{1}{(s^2-a^2)^{3/2}}$$
Im not familiar with laurent series please help me
 [/INST] Using 
$$(1+x)^n=1+nx+\frac{n(n-1)}{2!}x^2+\frac{n(n-1)(n-2)}{3!}x^3+\cdots$$
we have
\begin{eqnarray}
\frac{1}{\sqrt{(s^2-a^2)^3}}&=&(s^2-a^2)^{-\frac32}\\
&=&s^{-3}\left(1-\left(\dfrac{a}{s}\right)^2\right)^{-\frac32}\\
&=&\frac{1}{s^3}\left(1
-\left(-\dfrac{3}{2}\right)\left(\dfrac{a}{s}\right)^2
+\frac{(-\frac{3}{2})(-\frac{3}{2}-1)}{2!}\left(\dfrac{a}{s}\right)^4
-\frac{(-\frac{3}{2})(-\frac{3}{2}-1)(-\frac{3}{2}-2)}{3!}\left(\dfrac{a}{s}\right)^6
+\cdots\right)\\
&=&1
+\dfrac{3a^2}{2s^5}
+\dfrac{15a^4}{8s^7}
+\dfrac{35a^6}{16s^9}
+\cdots
\end{eqnarray}
</s>