A calculus problem by Swapnil Roge

Here is one in terms of double integrals:

$\large{I = \int_0^\infty e^{-\frac{x^2}{8}} dx \\ I^2 = \int_0^\infty e^{-\frac{x^2}{8}} dx \, \int_0^\infty e^{-\frac{y^2}{8}} dy \\ = \int_0^{\pi/2} \int_0^\infty e^{-\frac{r^2}{8}} \, r \, dr \, d\theta \\ = \frac{\pi}{2} \int_0^\infty e^{-\frac{r^2}{8}} \, r \, dr \\ = \frac{\pi}{2} \int_0^\infty e^{-u} \, r \, \frac{4}{r} \, du \\ = 2 \pi \int_0^\infty e^{-u} \, du = 2 \pi \\ I^2 = 2 \pi \implies \boxed{I = \sqrt{2 \pi}}}$

$\begin{aligned} I & = \int_0^\infty e^{-\frac {x^2}8} dx & \small \color{#3D99F6} \text{Let }t=\frac {x^2}8 \implies dt = \frac x4 \ dx \\ & = \int_0^\infty \sqrt 2 t^{-\frac 12} e^{-t} dt & \small \color{#3D99F6} \text{Gamma function }\Gamma (s) = \int_0^\infty t^{s-1} e^{-t} dt \\ & = \sqrt 2 \Gamma \left(\frac 12\right) \\ & = \sqrt{2\pi} \\ & \approx \boxed{2.507} \end{aligned}$

$Let\quad I=\int _{ 0 }^{ \infty }{ { e }^{ -\frac { { x }^{ 2 } }{ 8 } } } dx$

$substitute\quad \frac { { x }^{ 2 } }{ 8 } =t$

$\therefore x=\sqrt { 8t }$

$also,\quad 2\frac { x }{ 8 } dx=dt$

$\therefore dx=\frac { 4 }{ \sqrt { 8t } } dt=\sqrt { 2 } { \quad t }^{ -\frac { 1 }{ 2 } }dt$

$\therefore I=\int _{ 0 }^{ \infty }{ { e }^{ -t } } \sqrt { 2 } { t }^{ -\frac { 1 }{ 2 } }dt$

$\therefore I=\sqrt { 2 } \int _{ 0 }^{ \infty }{ { e }^{ -t } } { t }^{ -\frac { 1 }{ 2 } }dt$

$\therefore I=\sqrt { 2 } \int _{ 0 }^{ \infty }{ { e }^{ -t } } { t }^{ \left( \frac { 1 }{ 2 } -1 \right) }dt$

$but\int _{ 0 }^{ \infty }{ { e }^{ -t } } { t }^{ \left( \frac { 1 }{ 2 } -1 \right) }dt=\Gamma \left( \frac { 1 }{ 2 } \right) =\sqrt { \pi }$

$\therefore I=\sqrt { 2 } \sqrt { \pi } =\sqrt { 2\pi } =\boxed{2.507}$