A calculus problem by Rahil Sehgal

$\displaystyle \int_{-\infty}^{\infty} e^{-t^2} dt=\displaystyle 2 \int_{0}^{\infty} e^{-t^2} dt$ ....(Even function)

Put $x^2=z => 2xdx=dz$ .Substituting this into integral $2 \int_{0}^{\infty} e^{-z} .1/2.\sqrt{z} dz=\Gamma(1/2)=\sqrt{\pi}$

Now come to main integral $\int_{-\infty}^{\infty} e^{-x^2} dx \times \int_{-\infty}^{\infty} e^{-y^2} dx \times \int_{-\infty}^{\infty} e^{-z^2} dx \\ =(\sqrt{\pi})^3=\pi^{3/2}$

Using polar coordinates:

$\large \displaystyle \begin{cases} x = r \cdot \cos(\theta) \cdot \sin(\phi) \\ y = r \cdot \sin(\theta) \cdot \sin(\phi) \\ z = r\cdot \cos(\phi) \end{cases}$

$\large \displaystyle 0 < r < \infty, 0 < \theta < 2\pi, -\pi < \phi < \pi$

$\large \displaystyle |J| = r^2 \cdot \sin(\phi)$

The integral becomes:

$\large \displaystyle \int_0^{2\pi} \int_0^{\pi} \int_0^{\infty} r^2 \sin(\phi) e^{-r^2} dr d\phi d\theta$

$\large \displaystyle = \int_0^{2\pi} d\theta \int_0^{\pi} \sin(\phi) d\phi \int_0^{\infty} r^2 e^{-r^2} dr$

$\color{#20A900} \boxed {\large \displaystyle = 4 \pi \int_0^{\infty} r^2 e^{-r^2} dr}$

Now, let us define:

$\large \displaystyle I(k) = \int_0^{\infty} e^{-km^2} dm$

$\large \displaystyle I(k) = \frac12 \int_{-\infty}^{\infty} e^{-km^2} dm$

$\large \displaystyle I(k)^2 = \frac14 \int_{-\infty}^{\infty} \int_{-\infty}^{\infty} e^{-k\cdot(m^2+n^2)} dm dn$

Using polar coordinates, $m = \rho \cos(\alpha)$ , $n = \rho \sin(\alpha)$ , $|J| = \rho$ :

$\large \displaystyle I(k)^2 = \frac14 \int_0^{2 \pi} \int_0^{\infty} \rho e^{-k \rho^2} d\rho d\alpha$

$\large \displaystyle I(k)^2 = \frac14 \int_0^{2 \pi} d\alpha \int_0^{\infty} \rho e^{-k \rho^2} d\rho$

$\large \displaystyle I(k)^2 = \frac14 2\pi \cdot \frac{1}{2k} e^{-k\rho^2} \Bigg | _{\infty}^0$

$\large \displaystyle I(k)^2 = \frac{\pi}{4k}$

$\large \displaystyle I(k) = \sqrt{\frac{\pi}{4k}}$

$\color{#20A900} \boxed {\large \displaystyle \int_0^{\infty} e^{-km^2} dm = \frac12 \sqrt{\frac{\pi}{k}} }$

Differentiating with respect to $k$ in both sides:

$\large \displaystyle - \int_0^{\infty} m^2 e^{-km^2} dm = - \frac14 \sqrt{\frac{\pi}{k^3}}$

$\large \displaystyle \int_0^{\infty} m^2 e^{-km^2} dm = \frac14 \sqrt{\frac{\pi}{k^3}}$

Making $k = 1$ :

$\color{#20A900} \boxed { \large \displaystyle \int_0^{\infty} m^2 e^{-m^2} dm = \frac14 \sqrt{\pi} }$

Going back to our original integral:

$\large \displaystyle = 4 \pi \int_0^{\infty} r^2 e^{-r^2} dr$

$\large \displaystyle = 4 \pi \cdot \frac14 \sqrt{\pi}$

$\color{#20A900} \boxed { \large \displaystyle = \pi ^{\frac32} }$

So:

$\color{#3D99F6} a = 3, b = 2 , \boxed { \large \displaystyle a+b=5}$

$\displaystyle \int ^ { + \infty } _ { - \infty } \int ^ { + \infty } _ { - \infty } \int ^ { + \infty } _ { - \infty } e ^ { - x ^ { 2 } - y ^ { 2 } - z ^ { 2 } } dx dy dz = \pi ^ { \frac { 3 } { 2 } } = \pi ^ { \frac { a } { b } } \Rightarrow a = 3, b = 2 \Rightarrow a + b = 3 + 2 = \boxed { 5 }$