# Value of critical exponent $\alpha$ for 2D ising model

The Onsager solution for specific heat is

$$C\approx -Nk\frac{2}{\pi}\bigg(\frac{2J}{kT_c}\bigg)^2\ln\Big|1-\frac{T}{T_c} \Big|\qquad (T \textrm{ near } T_c)$$

Critical exponent $$\alpha\neq 0$$. Neither it follows a power law. But when we talk about scaling relation or the result we get for mean-field calculation, we say $$\alpha = 0$$.

Even when we plot any graph

(On x-axis temperature in units T$$\equiv Tk_B/J$$)

I cannot say $$C = \begin{cases} C_1 & \textrm{for TT_c} \end{cases}$$

where $$C_1,C_2$$ are some constants.

Let $$\tau=(T-T_{\rm c})/T_{\rm c}$$ and consider the behavior of some quantity $$X$$ as $$T$$ approaches $$T_{\rm c}$$ (that is, $$\tau$$ approaches $$0$$). Informally, if the quantity $$X$$ has a behavior of the type $$X \sim \tau^a$$ as, say, $$\tau\to 0^+$$ (convergence to $$T_{\rm c}$$ form above), then $$a$$ is the corresponding critical exponent. Now, what would be a precise definition of the exponent $$a$$? The standard way to define it is as $$a = \lim_{\tau\to 0^+} \frac{\log \lvert X\rvert}{\log \tau}. \tag{\star}$$ Check that, if $$X \sim \tau^a$$, then the right-hand side gives you indeed the correct exponent. (Note that often the exponent is negative; in this case, one usually considers its absolute value.)
Now, in the case of the specific heat of the 2d Ising model, $$C \sim \log\lvert\tau\rvert$$. Plugging this into ($$\star$$) indeed gives you the critical exponent $$\alpha = \lim_{\tau\to 0^+} \log\lvert\log\tau\rvert/\log\tau = 0$$. (The same is obviously also true when $$\tau\to 0^-$$.)