$\gamma,(\Gamma)$ -day Problem 1

$I=\int _{ 0 }^{ \infty }{ { t }^{ 7/2 }{ e }^{ -t/4 } } dt\\ \\ ={ 2 }^{ 9 }\int _{ 0 }^{ \infty }{ { x }^{ 7/2 }{ e }^{ -x } } dx\quad \quad \left( \cfrac { t }{ 4 } =x \right) \\ \\ ={ 2 }^{ 9 }\int _{ 0 }^{ \infty }{ { x }^{ \cfrac { 9 }{ 2 } -1 }{ e }^{ -x } } dx\\ \\ ={ 2 }^{ 9 }\Gamma \left( 9/2 \right) ={ 2 }^{ 9 }\Gamma \left( 4+\cfrac { 1 }{ 2 } \right) \\ \\ =\cfrac { { 2 }^{ 9 } }{ { 4 }^{ 4 } } \times \cfrac { 8! }{ 4! } \Gamma \left( \cfrac { 1 }{ 2 } \right) \quad \quad \quad \quad \quad \left( as\quad \Gamma \left( n+\cfrac { 1 }{ 2 } \right) =\cfrac { 2n! }{ { 4 }^{ n }n! } \Gamma \left( \cfrac { 1 }{ 2 } \right) \right) \\ \\ =5955.44\quad \quad \quad \quad (as\quad \Gamma \left( \cfrac { 1 }{ 2 } \right) =\sqrt { \pi } )$

We will use the following $\Gamma(x)=\int_0^{\infty}t^{x-1}e^{-t}\,dt$ $\Gamma(\frac{1}{2}+n)=\frac{(2n-1)!!}{2^n}\sqrt{\pi}$ The integral can be written as $I=\int_0^{\infty}t^{\frac{7}{2}}e^{-\frac{t}{2}}\,dt$ Take $y=\frac{t}{2}$ $\Rightarrow I=2^{\frac{7}{2}}\int_0^{\infty}y^{\frac{9}{2}-1}e^{-y}\,dt$ $\Rightarrow I=2^{\frac{7}{2}}\Gamma(\frac{9}{2})=2^{\frac{7}{2}}\Gamma(\frac{1}{2}+4)$ $\Rightarrow I=2^{\frac{1}{2}+3}\frac{(7)!!}{2^4}\sqrt{\pi}=\frac{105\sqrt{2\pi}}{2}=\boxed{131.59}$

Relevant wiki: Gamma Function

$\begin{aligned} I & = \int_0^\infty \sqrt{\frac {t^7}{e^\frac t2}} dt \\ & = \int_0^\infty t^\frac 72 e^{-\frac t4} \ dt & \small \color{#3D99F6} \text{Let }x = \frac t4 \implies 4 \ dx = dt \\ & = 2^9 \int_0^\infty x^\frac 72 e^{-x} \ dx & \small \color{#3D99F6} \text{Gamma function }\Gamma (s) = \int_0^\infty t^{s-1}e^{-t}\ dt \\ & = 2^9 \Gamma \left(\frac 92\right) & \small \color{#3D99F6} \text{Using }\Gamma(1+s) = s\Gamma(s) \text{ repeatedly} \\ & = 2^9 \cdot \frac 12 \cdot \frac 32 \cdot \frac 52 \cdot \frac 72 \cdot \color{#3D99F6} \Gamma \left(\frac 12\right) & \small \color{#3D99F6} \text{and }\Gamma \left(\frac 12\right) = \sqrt \pi \\ & = 3360 \color{#3D99F6} \sqrt \pi \\ & \approx \boxed{5955.445} \end{aligned}$