Question

$\color{red}\displaystyle \rm \int_{0}^{ \infty } \sum_{n = 0}^{ \infty } \frac{( - 1 {)}^{n} \: {x}^{2n + 1} }{ {2}^{n} \cdot n!} \sum_{n = 0}^{ \infty } \frac{ {x}^{2n} \: dx }{ {2}^{2n} \cdot(n! {)}^{2} }$​

Answer 1

$\small\color{red}\displaystyle \rm \int_{0}^{ \infty } \sum_{n = 0}^{ \infty } \frac{( - 1 {)}^{n} \: {x}^{2n + 1} }{ {2}^{n} \cdot n!} \sum_{n = 0}^{ \infty } \frac{ {x}^{2n} \: dx }{ {2}^{2n} \cdot(n! {)}^{2} }$

$\small\color{red}\displaystyle \rm \int_{0}^{ \infty } x {e}^{ \frac{ - {x}^{2} }{2} } \sum_{n = 0}^{ \infty } \frac{ {x}^{2n} }{ {2}^{2n} (n! {)}^{2} } \: dx$

$\small\color{red}\displaystyle \rm \int_{0}^{ \infty } x {e}^{ - u} \sum_{n = 0}^{ \infty } \frac{ {(2u)}^{n} }{ {2}^{2n} (n! {)}^{2} } \: \frac{du}{x}$

$\small\color{red}\displaystyle \rm \int_{0}^{ \infty } {e}^{ - u} \sum_{n = 0}^{ \infty } \frac{ {u}^{n} }{ {2}^{n} (n! {)}^{2} } \: du$

$\small\color{red}\displaystyle \rm \sum_{n = 0}^{ \infty } \frac{ 1 }{ {2}^{n} (n! {)}^{2} } \int _{0}^{ \infty } {e}^{ - u} \cdot {u}^{n} \: du$

$\small\color{red}\displaystyle \rm \sum_{n = 0}^{ \infty } \frac{ 1 }{ {2}^{n} (n! {)}^{2} } \: n!$

$\small\color{red}\displaystyle \rm \sum_{n = 0}^{ \infty } \frac{ 1 }{ {2}^{n} \cdot n! {} }$

$\small\color{red}\displaystyle \rm \sum_{n = 0}^{ \infty } \frac{ ( \frac{1}{2} {)}^{n} }{ n! {} }$

$\rm \red{ {e}^{ \frac{1}{2} } = \sqrt{e} }$