I know that:
$$\int_0^\infty \frac{1}{(x^2+1)^{3/2}}\,dx=1$$
On the other hand $$\int_0^\infty \frac{ dx}{({x^2+1^2)}^{3/2}}=\frac{(-1)^{3/2-1}\pi \Gamma (1/2)}{2\sin(\pi/2)(3/2-1)!\Gamma(1/2-3/2+1)}, \quad 0<3$$
This form of the equation is:
$$\frac{i\pi \sqrt \pi}{2\frac{\sqrt \pi}{2}\Gamma[0]}=1$$
Which implies that: $$\frac {i\pi}{\Gamma(0)}=1$$
in another word! Euler gamma function over zero is imaginary number times by pi number: $i\pi=\Gamma(0)$ which is a transcendental number.
Whether is my conclusion about Euler Gamma function correct? is it new conclusion?
$$\int_0^\infty \frac{x^m \, dx}{({x^n+a^n)}^r}=\frac{(-1)^{r-1}\pi a^{m+1-nr}\Gamma [(m+1)/n]}{n\sin[\pi (m+1)/n](r-1)!\Gamma[((m+1)/n)-r+1]}, \quad 0<m+1<nr$$
set $m=0$, $a=1$, $n=2$ and $r=3/2$
Note that the formula for this integral (from your other question here):
$$\int_0^\infty \frac{x^m \, dx}{({x^n+a^n)}^r}=\frac{(-1)^{r-1}\pi a^{m+1-nr}\Gamma [(m+1)/n]}{n\sin[\pi (m+1)/n](r-1)!\Gamma[(m+1)/n-r+1]}, \quad 0<m+1<nr$$
In terms of the Gamma Function is only valid for integer values of $r$ (GEdgar points out the same thing in the comments).
Your conclusion is obviously contradictory anyway, $\Gamma(0) = \tilde{\infty} \neq i\pi$.