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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 $r=3/2$

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$\Gamma(0) = \tilde{\infty}$. – George V. Williams Feb 7 '13 at 22:50
You must have made a mistake somewhere since the Gamma function "blows up" at the origin. – Pedro Tamaroff Feb 7 '13 at 23:10
@Peter Tamaroff maybe this point that in terms of the Gamma Function is only valid for integer values – Neo Feb 7 '13 at 23:27
$(3/2-1)!$ is not defined. – Yves Daoust Jun 12 '14 at 12:24

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, $\Gamma(0) = \tilde{\infty} \neq i\pi$.

share|cite|improve this answer
What is infinity with a tilde over it? – Gerry Myerson Feb 7 '13 at 23:20
@GerryMyerson, Complex infinity (infinity in the complex plane with an undefined argument). I get my notation from Wolfram Functions Site. – George V. Williams Feb 7 '13 at 23:26
Thanks. Never saw it before. – Gerry Myerson Feb 7 '13 at 23:53

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