Klein Gordon Equation from Correspondence Principle in Relativistic domain

The Klein Gordon equation is developed from the general relativistic energy momentum relation by the substitution of corresponding operators as, $$ E = i\hbar\frac{\partial}{\partial{t}}$$ $$ \vec{p} = -i\hbar\nabla$$
acting on $\phi(x,t)$ to give, 
$$ -\hbar^2\frac{\partial^2\phi}{\partial{t^2}} = -\hbar^2c^2\nabla^2\phi +m^2c^4\phi$$ $\rightarrow$$$\hbar^2c^2\left[\nabla^2\phi-\frac{1}{c^2}\frac{\partial^2\phi}{\partial{t^2}} \right]= m^2c^4\phi$$ $\rightarrow$$$\left[\Box+\frac{m^2c^2}{\hbar^2}\right]\phi(x_i)=0$$
where the d'alembertian operator is defined here as $$ \Box =  \frac{1}{c^2}\partial_t^2 - \nabla^2$$

We know that, $$p^\mu{p_\mu} = -\hbar^2 \partial^\mu\partial_\mu = -\hbar^2\Box$$ where the minkowski metric is given by,$ \eta_{\mu\nu}$ = diagonal(1,-1,-1,-1), and thus we have,
$$ \left(\partial^\mu\partial_\mu + \frac{m^2c^2}{\hbar^2}\right)\phi = 0 $$ which is known as Klein Gordon Equation.
To make life simple, we adopt to the convention,
where we equate the planck's constant and the speed of light equal to 1 (dimensionless number). The consequences are, the dimension of Length and Time are the same and the dimension of Mass is just the inverse of Length or Time. In addition, Energy and Momentum is measured in the same unit as of the Mass. 

And, contravariant vectors are $ A^{\mu}= (A_0, \vec{A})$ and corresponding covariant transformation is $ A_{\mu}= \eta_{\mu\nu}A^{\nu} = (A_0, -\vec{A})$ where the differential is defined in reverse way 
$ \partial_\mu = \left(\partial_0,\vec{\nabla}\right) $
[$\partial_0 = \partial_t$] and 
$ \partial^{\mu}= \eta^{\mu\nu}\partial_{\nu} = ({\partial}_0,-\vec{\nabla})$
With these substitution we get our KG equation as, $$\left(\Box + m^2\right)\phi = 0 $$ where $p^\mu{p}_\mu = m^2 \\ \,\, E^2 = \omega^2 = m^2 + {|\vec{p}|^2} $
Since $m^2$ and d'Alembertian operator is invariant under Lorentz transformation, if the function $\phi$ satisfies the condition $ \phi'(x') = \phi(x)$ then the whole KG equation is invariant under Lorentz transformation. 
But, there are some problems with this equation in the basic definition of $\phi(x)$. First of all it cannot be the wave function of the particle as it is in the case of Non-relativistic Schr\"{o}dinger equation.

The problem arises because of the probability statements. To understand its significance, let us just assume that $\phi(x)$ is the usual wave function. 

Then, it says that the probability of finding the particle at a position x is the same as of the finding the particle in $x'$ position in some other reference frame [Lorentz invariance condition]. It implies the wave function should behave like a scalar quantity and independent of direction, which is not true in general for spin half particles. The properties of spin half particles depends on from which direction it is measured and change with respect to different orientations of the reference frame. 

And the second and the most important problem is that the probability density definition changes completely and it allows for the weird possibility of the negative probability. 
The continuity equation with the time component becomes, $$\partial_0\rho+\vec{\nabla}\cdot\vec{J} = 0 = \partial_\mu{J}^\mu$$ where $ \rho = \frac{1}{2} \left[\phi^*(\partial_0\phi) - (\partial_0\phi)^*\phi\right] = \frac{1}{2} \phi^*\overleftrightarrow{\partial_0}\phi $ 
and \newline
$\vec{J} = \frac{-1}{2}\left[\phi^*(\nabla\phi) - (\nabla\phi)^*\phi\right] = \frac{1}{2} \phi^*\overleftrightarrow{\partial_0}\phi $
where $\rho$ is the equivalent probability density as defined in non-relativistic case can now have positive as well as negative values, that is not compatible with the definition of probability - you can check it for monochromatic wave of the for $\phi = Ae^{\pm{ikx}}$. Thus, either we will have to abandon KG equation or should find an alternative way of description for its definition.

Fourier series and Fourier transform

The concept of Fourier series is based on the orthogonality of sine and cosine functions. So, it is a perfect thing to start from these properties.
A sine function with time period $2\pi$ radians is given by $sin(t)$ and the integral of the function over its time period is zero. In general, for an arbitrary sine function with Time period $T_0$, the function is given by $sin (\frac{2n\pi{t}}{T_0})$, where the time integral over this new time period is zero. Everything applies to "cosine" function as much as the same.

In functional form, $$sin\left(\frac{2n\pi}{T_0}(t+T_0)\right) = sin(\frac{2n\pi{t}}{T_0})$$ and $$\int_{t_0}^{t_0+T_0} \,dt sin(\frac{2n\pi{t}}{T_0}) = \int_{t_0}^{t_0+T_0} \,dt cos(\frac{2m\pi{t}}{T_0}) = 0  \\ for \,\,\,\,n,m > 0 $$
From our trigonometric identities, we can write $$ sinA cosB = \frac{sin(A+B)+ sin(A-B)}{2}$$ Using this in our case, we will get (n+m) and (n-m) where n and m are integers. This gives, $$\int_{t_0}^{t_0+T_0} \,dt sin(\frac{2n\pi{t}}{T_0}) cos(\frac{2m\pi{t}}{T_0}) = 0 $$
Similarly, with the other identity $$ sinA sinB = cos (A-B) - cos(A+B) \\ cosA cosB = cos(A+B) + cos(A-B)$$ we can prove $$\int_{t_0}^{t_0+T_0} \,dt sin(\frac{2n\pi{t}}{T_0}) cos(\frac{2m\pi{t}}{T_0}) = 0 $$ except now, there arise a problem for the case n = m as we will get an extra non-zero term $$\int_{t_0}^{t_0+T_0} \,dt \frac{cos(\frac{2(n-m)\pi{t}}{T_0})}{2} = \int_{t_0}^{t_0+T_0} \frac{1}{2}\,dt = \frac{T_0}{2}$$
Finally we get the following identities,
$$\int_{t_0}^{t_0+T_0} \,dt sin(\frac{2n\pi{t}}{T_0}) cos(\frac{2m\pi{t}}{T_0}) = 0 $$ and
$$\int_{t_0}^{t_0+T_0} \,dt sin(\frac{2n\pi{t}}{T_0}) sin(\frac{2m\pi{t}}{T_0}) = \int_{t_0}^{t_0+T_0} \,dt cos(\frac{2n\pi{t}}{T_0}) cos(\frac{2m\pi{t}}{T_0}) = \delta_{nm}\frac{T_0}{2} $$
This is the fundamental orthogonality relation we require for the sine and cosine function to serve as the basis vectors.
Now, the Fourier series is defined as, $$f(t) = \frac{a_0}{2} + \sum_{n=1}^\infty a_n cos(\frac{2n\pi{t}}{T_0}) + \sum_{n=1}^\infty b_n sin(\frac{2n\pi{t}}{T_0})$$
The only unknown terms in this series are the Fourier coefficients $a_0, a_n, b_n$ which can be determined using our orthogonality relations as, $$ a_n = \frac{2}{T_0}\int_{t_0}^{t_0+T_0} \,dt cos(\frac{2n\pi{t}}{T_0}) f(t) \,\,\,\, n = 0,1,2..\\ b_n = \frac{2}{T_0}\int_{t_0}^{t_0+T_0} \,dt sin(\frac{2n\pi{t}}{T_0}) f(t) $$
Once we are able to write a function in terms of sine and cosine, we proceed to write it in terms of exponential functions using Euler's formula. And we can contract the series in a simple form as, $$ f(t) = \sum_{n=-\infty}^\infty c_n e^{i\frac{2n\pi{t}}{T_0}}$$
Now, the only unknown in this series is $c_n$ and it can also be determined using the orthogonality relation of exponentials $$ \int_{t_0}^{t_0+T_0}\,dt\,e^{i\frac{2n\pi{t}}{T_0}}e^{-i\frac{2m\pi{t}}{T_0}} = \delta_{nm}T_0 $$ which gives finally, $$ c_m = \frac{1}{T_0} \int_{t_0}^{t_0+T_0} f(t) e^{\frac{i2n\pi{t}}{T_0}} \,dt $$

Fourier Transform:

Substituting for $ \omega_n = \frac{2n\pi}{T_0}$ we get $$ f(t)  = \sum_{n= -\infty}^\infty c_n e^{i\omega_nt}$$ and $$ c_m = \frac{1}{T_0} \int_{-\frac{T_0}{2}}^{\frac{T_0}{2}} f(t) e^{-i\omega_nt}\,dt$$ (from the substitution $t_0 = -\frac{T_0}{2}$)

It is easily seen from the above equation that, it fails for $T_0 = \infty$ For a function with infinite period, the coefficient cannot be calculated from above equation.

Since, many of the real function used in Physics has this structure (Time period infinity), it is necessary to look our for an alternative to deal with these functions. This is where the integral comes about to play the significant role and known as fourier transform.

For very large time period, the change in $\omega$ - the step size becomes negligible i.e. $$\omega_{n+1} -\omega_n = \frac{2\pi}{T_0} \\ \rightarrow \,\,\,\, \delta\omega = d\omega\\ \frac{1}{T_0} = \frac{d\omega}{2\pi}$$ and there is no need for the index 'n'.

Combining everything together, $T_0\rightarrow\infty$ and substituting for $c_n$ with some dummy variable, $$ f(t) = \sum_{n= -\infty}^\infty\frac{d\omega}{2\pi} \left[\int_{-\infty}^{\infty}f(\tau) e^{-i\omega_n\tau}\,d\tau\right]e^{i\omega_n{t}}$$ As "n" changes $d\omega$ is very small and the sum can be replaced by Integral,

$$ f(t) = \frac{1}{2\pi}\int_{-\infty}^{\infty} e^{i\omega_nt}\,d\omega \left[\int_{-\infty}^\infty f{\tau} e^{-i\omega_n\tau}\,d\tau\right] $$ the bracketed term depends only on $\omega$. Thus, we have finally arrived at our fourier and inverse fourier transform. It doesn't matter which one you call the fourier transform and which one is the inverse one. By splitting $\frac{1}{2\pi}$ factor in two different ways, the final equations are, $$f(t) = \frac{1}{\sqrt{2\pi}} \int_{-\infty}^{\infty}\,d\omega e^{i\omega{t}}F(\omega) \\ F(\omega) = \frac{1}{\sqrt{2\pi}} \int_{-\infty}^{\infty}\,dt \,e^{-i\omega{t}}\,f(t) \,\,\,\,\\or\,\,\,\,\\ f(t) = \frac{1}{2\pi} \int_{-\infty}^{\infty}\,d\omega e^{i\omega{t}}F(\omega) \\ F(\omega) = \int_{-\infty}^{\infty}\,dt e^{-i\omega{t}}\,f(t)$$ or you can put the $2\pi$ factor in the other way. 

Volume of a Hypersphere in N - dimension

    To find the number of micro states in a statistical system for N number of particles in a fixed volume and energy within the range E and $E+\delta{E}$, we require the knowledge on how to find the volume for N dimensional hypersphere in momentum space.

In the simplest way, the problem is to find the volume of an N-dimensional hypersphere. If we try to find the general formula by mathematical induction, using the equation in Cartesian form, it will produce a complex equation.

    Instead there is a clever trick to simplify the procedure based on the concept of infinity. The equation of a sphere in Cartesian form is $$ x_1^2+x_2^2+x_3^2+...+x_N^2 = r^2$$ where 'r' is the radius of the sphere.

    In general, volume of an n-dimensional sphere should be proportional to $r^N$ and its surface is proportional to $r^{N-1}$. So, $$V \propto r^N\\ S \propto r^{N-1} $$ or $$ S = A r^{N-1}$$ where A is the proportionality constant that is equal to the surface area of a unit sphere. If we could find somehow this value of A in n-dimension, it is then a simple task to proceed to the volume. The special characteristic of the surface of a sphere is that, once we know the surface, the volume is just a simple line integral of the surface value along 'r'.

$$ dV = S dr  = A r^{N-1} dr $$

The generalization of the equation of sphere in Cartesian form is a straight task. To connect all these things, we use the property of the infinity given as,

    Integral of a function over the infinite volume is the same independent of whether the integral is done with Cartesian cubic volume element (cubic volume extending to infinity) or spherical volume element (spherical volume extending to infinity), the result is the same.  This is not a completely new one because we use this same fact to find the integral of $$\int_{-\infty}^{\infty}e^{-\alpha{x^2}}\,dx = \sqrt{\frac{\pi}{\alpha}}$$

    Using the same integral we have, $$ I = \int_{-\infty}^{\infty} e^{-{x_1^2+x_2^2+...+x_N^2}} \,dx_1\,dx_2\,...\,dx_N = \pi^{N/2} $$ Now,the same function integrated over spherical volume element gives, $$ I = \int_0^{\infty} e^{-{x_1^2+x_2^2+...+x_N^2}} A r^{N-1}dr = \int_0^{\infty} A e^{-{r^2}} r^{N-1} dr $$ Substituting $y = r^2 $ we have $$I  = \int_0^{\infty} A e^{-y} y^{\frac{N-1}{2}} \,\frac{1}{2y^{1/2}}\,dy = \int_0^{\infty} A \frac{1}{2} e^{-y} y^{\frac{N}{2}-1}dy = \frac{1}{2} A \Gamma(N/2)$$ then we have, by equating the value of I in two ways, $$ A = \frac{2\pi^{N/2}}{\Gamma(N/2)}$$ Now, the volume is evaluated to be $$V  =  \int _0^R \,S \,dr  = \int_0^R \,A r^{N-1}\,dr \\ V =  \int_0^R\,\frac{2\pi^{N/2}}{\Gamma(N/2)} r^{N-1} \, dr = \frac{\pi^{N/2}}{\Gamma(N/2)} \frac{R^{N}}{N/2} = \frac{\pi^{N/2} R^N}{\Gamma(\frac{N}{2}+1)}$$ where we used the fact  $$\frac{N}{2} \Gamma(\frac{N}{2}) = \Gamma(\frac{N}{2}+1)$$

Thus we can determine the volume and so the surface area of an n-dimensional hypersphere in a simple form.