What is $\Gamma(1/4)$? I'm trying to figure out how you would go about solving it

Question

We know that $$\Gamma(1/2) = \int e^{-x} x^{-1/2} \text{d}x$$

If I set $t^2 = x$ then we have $2\int e^{-t^2} \text{d}t$

We know that

$$\frac{1}{\sqrt{\pi}}\int e^{-t^2} \text{d}x = \frac{1}{2}$$

This is simply half of the area under the graph of a normal distribution given that my parameters of integration are $(0, +\infty)$.

Hence $$2\int e^{-t^2} \text{d}x = 2 \left(\frac{1}{2}\sqrt{\pi}\right) = \sqrt{\pi}$$

Setting up the problem similarly for $\Gamma(1/4)$ would give us

$$\int e^{-x} x^{-3/4} \text{d}x$$

Then setting $t^4 = x$ gives us $$4\int e^{-t^4} \text{d}t$$

Not exactly sure where to go from here or if this even the right approach. Any help is appreciated. Thank you so much.

Answer 1

Well, $\Gamma(1/4)$ is transcendental; in fact it is algebraically independent of $\pi$. It is not known to be expressible in terms of more basic numbers. In fact, lots of answers are written in terms of $\Gamma(1/4)$.

SEE

Answer 2

This is about the integral definition of Euler Gamma Function:

$$\Gamma(x) = \int_0^{+\infty} t^{x-1} e^{-t}\ \text{d}t$$

Since in your case $x = 1/4$ we have

$$\Gamma(1/4) = \int_0^{+\infty} t^{-3/4}e^{-t}\ \text{d}t$$

You may perform the change of variable you like the most, but you won't get a suitable answer, if you're looking for numbers. Indeed, $\Gamma(1/4)$ is a transcendental number whose value is ish

$$\Gamma\left(\frac{1}{4}\right) \approx 3.62561(...)$$

There is only a "way" to express it in other terms, which will be circular since it will involve another transcendental number. Anyway, by Legendre duplication formula we have

$$\Gamma(z)\Gamma\left(z + \frac{1}{2}\right) = 2^{1-2z}\sqrt{\pi}\ \Gamma(2z)$$

Plugging $z = 1/4$ and isolating the first term:

$$\Gamma\left(\frac{1}{4}\right) = \frac{2^{1-1/2}\sqrt{\pi}\Gamma\left(\frac{1}{2}\right)}{\Gamma\left(\frac{3}{4}\right)}$$

Now we know $\Gamma\left(\frac{1}{2}\right) = \sqrt{\pi}$ hence

$$\Gamma\left(\frac{1}{4}\right) = \frac{\pi\sqrt{2}}{\Gamma\left(\frac{3}{4}\right)}$$

And as you see, another transcendental number appeared.

Answer 3

Since another answer gives $\Gamma\left(\frac14\right)$ in other contexts, here are some constants where it appears in:

Leminscate constant$=L$:

$$\Gamma\left(\frac14\right)=2^\frac34\sqrt[4]\pi\sqrt L$$

Gauss Constant$=G$:

$$\Gamma\left(\frac14\right)=(2\pi)^\frac34\sqrt G$$

Second leminscate Constant$=L_2$:

$$\Gamma\left(\frac14\right)=\frac{\sqrt[4]2\pi^\frac34}{\sqrt{L_2}}$$

There are other constants, but these are some major ones.