How does $\int_0^\infty e^{-t^4}dt = \Gamma (\frac{5}{4}) ?$

Question

My text book claims that $$\int_0^\infty e^{-t^4}dt = \Gamma \left(\frac{5}{4}\right).$$ I fail to see this. By the definition of the gamma function we have $$\Gamma (z) = \int_0^\infty t^{z-1}e^{-t}dt.$$ Thus, plugging in $\frac{5}{4}$ for $z$ should give us $$\begin{array}{ccc} \Gamma\left(\frac{5}{4}\right) & = & \int_0^\infty t^{\frac{5}{4}-1}e^{-t}dt \\ & = & \int_0^\infty t^{1/4}e^{-t}dt. \\ \end{array}$$ This algebraically implies that $$t^{1/4}e^{-t} = e^{-t^4} \Leftrightarrow t^{1/4} = e^{-t^3}.$$ And from here I just get into a messy algebra scenario using logarithms that just really doesn't get me anywhere (as far as solving for $t$ goes).

So what am I missing here?

Answer 1

Let $u=t^4$ so $t=u^{1/4}$ and $dt=\frac 14 u^{-3/4}$ and then

$$\int_0^\infty e^{-t^4}dt=\frac14\int_0^\infty u^{-3/4}e^{-u}du=\frac14\int_0^\infty u^{-3/4}e^{-u}du=\frac14\Gamma\left(\frac14\right)=\Gamma\left(\frac54\right)$$ using the equality

$$\Gamma(x+1)=x\Gamma(x)$$

Answer 2

In general for $n > 0$, $$I_n = \int_0^{\infty} \exp(-t^n) dt = \Gamma(1+1/n)$$ This follows immediately from the substitution $x=t^n$. We then have $$I_n = \int_0^{\infty} \exp(-x) \dfrac{dx}{nt^{n-1}} = \int_0^{\infty} \exp(-x) \dfrac{dx}{nx^{1-1/n}} = \dfrac{\Gamma(1/n)}{n} = \Gamma(1+1/n)$$