# solve $T(n)=2T(\dfrac{n}{2})+\dfrac{8}{9}T(\dfrac{3n}{4})+\Theta(\dfrac{n^2}{\log{n}})$ using Akra-Bazzi method

Assume we have this recurrence: $$T(n)=2T(\dfrac{n}{2})+\dfrac{8}{9}T(\dfrac{3n}{4})+\Theta(\dfrac{n^2}{\log{n}})$$ We want to solve it using Akra-Bazzi method. As we know, $$\sum_{i=1}^k\dfrac{a_i}{b_i^p}=1$$. So $$\dfrac{2}{2^p}+\dfrac{\dfrac{8}{9}}{(\dfrac{4}{3})^p}=1$$ and we get $$p=2$$. We know: $$T(n)=\Theta\left(n^p(1+\int_1^n{\dfrac{f(x)}{x^{p+1}}\text{d}x})\right)$$ Which $$f(x)$$ is the cost of dividing and merging of the problem. Here, $$f(x)=\Theta(\dfrac{n^2}{\log{n}})$$. Using Akra-Bazzi method, integrations part converts to: $$\log{\log{x}}|_1^n$$. But for $$x=1$$ the integral is divergent. So, how can I deal with these kinds of problem that integral goes to infinity?

If you pick any $$n_1 > 1$$, $$\log \log n_1$$ exists and is just some constant. Then $$\int_{n_1}^n \frac{1}{x \log x} \text{d}x = \log \log n - \log \log n_1 = \Theta(\log \log n),$$ and the solution of the recurrence is $$\Theta(n^2 \log \log n)$$. For convenience you can pick $$n_1 = e$$.
Alternatively, you can rewrite your recurrence replacing $$f(n)=\frac{n^2}{\log n}$$ with some function $$g(n) = \Theta(f(n))$$ that yields an integrand that is defined over the domain of integration. E.g., $$g(n) = \frac{(2n)^2}{\log 2n}$$ results in the integral \begin{align*} \int_{1}^n \frac{4x^2}{x^3 \log 2x} \text{d}x&= \int_{1}^n \frac{4}{x \log 2x} \text{d}x = \int_{2}^{2n} \frac{4}{t \log t} \text{d}t = \left. 4 \log \log 2t\right|_{t=2}^{2n} \\ &= 4 \log \log 2n - 4 \log \log 2 = \Theta(\log \log n). \end{align*}