Triple Integral!

Since, the variables are independent to each other. We can write that as

$\begin{aligned} \int_{0}^{\infty} \int_{0}^{\infty} \int_{0}^{\infty} e^{-(x^2 + y^2 +z^2)} dz dy dx &= \int_{0}^{\infty} e^{-x^2} \left( \int_{0}^{\infty} e^{-y^2} \left(\int_{0}^{\infty} e^{-z^2} dz \right) dy \right) dx \\ &= \int_{0}^{\infty} e^{-x^2} dx \times \int_{0}^{\infty} e^{-y^2} dy \times \int_{0}^{\infty} e^{-z^2} dz \\ &= \left(\int_{0}^{\infty} e^{-z^2} dz \right)^3 \end{aligned}$

Take $z = t^{1/2} \iff dz = \dfrac{1}{2} t^{-1/2} dt$ . And that we get

$\int_{0}^{\infty} e^{-z^2} dz = \dfrac{1}{2} \int_{0}^{\infty} e^{-t} t^{-1/2} dt = \dfrac{1}{2} \Gamma (1/2) = \dfrac{\sqrt{\pi}}{2}$

So, the given integral evaluates to be $\boxed{\dfrac{\pi \sqrt{\pi}}{8}}$ .

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• $\Gamma(x)$ is gamma function, which is equal to $\Gamma(x) = \int_{0}^{\infty} e^{-t} t^{x-1} dt$ .

• And using reflection formula for Gamma function i.e. $\Gamma(z) \Gamma(1-z) = \dfrac{\pi}{\sin (\pi z)}$ , we get $\Gamma(1/2) = \sqrt{\pi}$ by taking $z= 1/2$ .