Logs and graphs

$\log(y)+\log^2(x) = 1 \implies y= e^{1-\log^2(x)}$

$y=0$ when it touches $x$ axis . So there are two limits $x\rightarrow{∞}$ and $x\rightarrow{0}$

Area = $\displaystyle\int_0^∞ e^{1-\log^2(x)} dx$ Take $\log(x)=t$ , the integral becomes= $\displaystyle e\int_{-∞}^{∞} e^{t-t^2} dt$ $= e\int_{-∞}^∞ e^{-(t-\frac{1}{2})^2+\frac{1}{4}}dt$ $= e^{\frac{5}{4}} \int_{-∞}^∞ e^{-(t-\frac{1}{2})^2} dt = e^{\frac{5}{4}} \sqrt{π} \textrm{(Gaussian Integral)}$ So $A= e^{\frac{5}{4}} \sqrt{π}$ and $A^4 = e^5 π^2 = (2^2+2)e^{2^2+1}\dfrac{π^2}{6}$

So $A^4 = (2^2+2) e^{2^2+1} \zeta(2)$ - Answer $\boxed{b=2}$

To be honest, it's rather easy to guess $b=2$ . Maybe you can give it like: $A^4=ae^b\zeta(c)$ And ask $a+b+c=?$ . What do you think, @Zakir Husain ?