$$T(n)=2T\left(\frac{n}{2}\right)+\frac{n}{\lg n}$$
For Master theorem, $a = 2, b = 2, n^{\log _b a} = n$. $f(n) = \frac{n}{\lg n}$. We cannot say $f(n) = O(n^{\log_b a - \epsilon})$, as the difference between $n$ and $\frac{n}{\lg n}$ is not polynomial. So, we cannot apply Master theorem. So, trying substitution. Since, we have a lg term, we can try all powers of 2.
$T(1) = 1$, assuming.
$T(2) = 2T(1) + \frac{2}{\lg 2} = 4$
$T(4) = 8 + 2 = 10$
$T(8) = 20 + \frac{8}{3} = 22.66$
$T(16) = 45.3 + 4 = 49.3$
$T(32) = 98.6 + 6.4 = 105$
$T(64) = 210 + 10.6 = 220.6$
Not able to reach a conclusion. We can see that the recurrence is between case 1 and case 2 of Master theorem. So, it is $\Omega \left(n\right)$ and $O\left(n \lg n\right)$. So, lets solve the recurrence directly.
$T(n)=2T\left(\frac{n}{2}\right)+\frac{n}{\lg n}$
$= 2^2 T\left(\frac{n}{4}\right)+\frac{n}{\lg n - 1} + \frac{n}{\lg n}$
$\dots$
$= 2^{\lg n} T(1) + \frac{n}{1} + \frac{n}{2} + \dots + \frac{n}{\lg n}$
$= n + n \left( \frac{1}{1} + \frac{1}{2} + \dots + \frac{1}{\lg n}\right)$
$=n + n \left(\lg \lg n + \gamma \right)\\= \Theta\left(n \lg \lg n\right)$
The sum of $\lg n$ terms in HP approximated to $\lg \lg n +\gamma$ where $\gamma$ is Euler Mascheroni constant.
Ref: https://en.wikipedia.org/wiki/Euler%E2%80%93Mascheroni_constant
