Question

$\huge{ \boxed{ \bold \red{ \int \limits_0^{1} {x}^{ - x} \: dx}}}$

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Answer 1

${ \displaystyle{\bold \red{ \int \limits_0^{1} {x}^{ - x} \: dx}}}$

${ \displaystyle{\bold \red{ \int \limits_0^{1} {e}^{ - x \: Inx} dx}}}$

${ \displaystyle{\bold \red{e {}^{x} = \sum \limits_ {n = 0} ^{ \infty} \frac{ {x}^{n} }{ {n}{ᴉ } } }}}$

${ \displaystyle{\bold \red{ \int \limits_0^{1} \sum \limits_ {n = 0} ^{ \infty} \frac{( - Inx) {}^{n} {x}^{n} }{nᴉ} \: dx }}}$

${ \displaystyle{\bold \red{ \sum \limits_ {n = 0} ^{ \infty} \int \limits_0^{1} \frac{( - Inx) {}^{n} {x}^{n}}{nᴉ} \: dx}}}$

$\large \bold \red{u=-Inx}$

$\large \bold \red{x = {e}^{ - u}}$

$\large \bold \red{dx = { - e}^{ - u} du}$

${ \displaystyle{\bold \red{ \sum \limits_ {n = 0} ^{ \infty} \frac{1}{nᴉ} \int \limits_0^{ \infty } u {}^{n} e {}^{ - (n + 1)u}du }}}$

$\bold \red{v =n + 1}$

$\bold \red{dv =du}$

${ \displaystyle{\bold \red{ \sum \limits_ {n = 0} ^{ \infty} \frac{1}{nᴉ} \int \limits_0^{ \infty } \frac{ {v}^{n} }{n + 1} e {}^{2 - v} \frac{dv}{n + 1} }}}$

${ \displaystyle{\bold \red{ \sum \limits_ {n = 0} ^{ \infty} \frac{1}{nᴉ} \frac{1}{(n + 1) {}^{n + 1} } \int \limits_0^{ \infty }v {}^{n - v}e \: dv }}}$

${ \displaystyle{\bold \red{ \sum \limits_ {n = 0} ^{ \infty} \frac{1}{nᴉ} \frac{1}{(n + 1) {}^{n + 1} } nᴉ }}}$

${ \displaystyle{\bold \red{ \sum \limits_ {n = 0} ^{ \infty} \cancel\frac{1}{nᴉ} \frac{1}{(n + 1) {}^{n + 1} } \cancel{nᴉ} = \sum \limits _{n = 1}^{ \infty } \frac{1}{ {n}^{n} } }}}$

${ \displaystyle{\bold \red{ \int \limits_0^{1} {x}^{ - x} \: dx =\sum \limits _{n = 1}^{ \infty } {n}^{ - n} }}}$

Answer 2

Answer:

${ \displaystyle{\bold \red{ \int \limits_0^{1} {x}^{ - x} \: dx}}}$

${ \displaystyle{\bold \red{ \int \limits_0^{1} {e}^{ - x \: Inx} dx}}}$

${ \displaystyle{\bold \red{e {}^{x} = \sum \limits_ {n = 0} ^{ \infty} \frac{ {x}^{n} }{ {n}{ᴉ } } }}}$

${ \displaystyle{\bold \red{ \int \limits_0^{1} \sum \limits_ {n = 0} ^{ \infty} \frac{( - Inx) {}^{n} {x}^{n} }{nᴉ} \: dx }}}$

${ \displaystyle{\bold \red{ \sum \limits_ {n = 0} ^{ \infty} \int \limits_0^{1} \frac{( - Inx) {}^{n} {x}^{n}}{nᴉ} \: dx}}}$

$\large \bold \red{u=-Inx}$

$\large \bold \red{x = {e}^{ - u}}$

$\large \bold \red{dx = { - e}^{ - u} du}$

${ \displaystyle{\bold \red{ \sum \limits_ {n = 0} ^{ \infty} \frac{1}{nᴉ} \int \limits_0^{ \infty } u {}^{n} e {}^{ - (n + 1)u}du }}}$

$\bold \red{v =n + 1}$

$\bold \red{dv =du}$

${ \displaystyle{\bold \red{ \sum \limits_ {n = 0} ^{ \infty} \frac{1}{nᴉ} \int \limits_0^{ \infty } \frac{ {v}^{n} }{n + 1} e {}^{2 - v} \frac{dv}{n + 1} }}}$

${ \displaystyle{\bold \red{ \sum \limits_ {n = 0} ^{ \infty} \frac{1}{nᴉ} \frac{1}{(n + 1) {}^{n + 1} } \int \limits_0^{ \infty }v {}^{n - v}e \: dv }}}$

${ \displaystyle{\bold \red{ \sum \limits_ {n = 0} ^{ \infty} \frac{1}{nᴉ} \frac{1}{(n + 1) {}^{n + 1} } nᴉ }}}$

${ \displaystyle{\bold \red{ \sum \limits_ {n = 0} ^{ \infty} \cancel\frac{1}{nᴉ} \frac{1}{(n + 1) {}^{n + 1} } \cancel{nᴉ} = \sum \limits _{n = 1}^{ \infty } \frac{1}{ {n}^{n} } }}}$

${ \displaystyle{\bold \red{ \int \limits_0^{1} {x}^{ - x} \: dx =\sum \limits _{n = 1}^{ \infty } {n}^{ - n} }}}$