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Question Number 113589 by mathdave last updated on 14/Sep/20

p r o v e t h e f o l l o w i n g i n t e g r a l

\int_{0}^{\frac{π}{2}} \frac{x^{2}}{\sin x} d x = 2 π G - \frac{7}{2} ζ (3)

\int_{0}^{1} \ln {[\frac{x + \sqrt{1 - x^{2}}}{x - \sqrt{1 - x^{2}}}]}^{2} \frac{x}{1 - x^{2}} d x = \frac{. π^{2}}{2}

\int_{0}^{\infty} \frac{e^{x} \ln x}{1 + e^{2 x}} d x = \frac{π}{2} \ln [\frac{Γ (\frac{3}{4})}{Γ (\frac{1}{4})} \sqrt{2 π}]

Answered by maths mind last updated on 15/Sep/20

\int_{0}^{\frac{π}{2}} \frac{x^{2}}{s i n (x)} d x = \int_{0}^{\frac{π}{2}} \frac{x^{2}}{2 t g (\frac{x}{2})} (1 + {t g}^{2} (\frac{x}{2})) d x = \int_{0}^{\frac{π}{2}} x^{2} \frac{d (t g (\frac{x}{2}))}{t g (\frac{x}{2})}

I B P = {[x^{2} l n (t g (\frac{x}{2}))]}_{0}^{\frac{π}{2}} - \int_{0}^{\frac{π}{2}} 2 x l n (t g (\frac{x}{2})) d x

= - 2 \int_{0}^{\frac{π}{2}} x l n (t g (\frac{x}{2})) d x

w e u s e f o u r i e r s e r i e o f l n (t g (x)), x \in] 0, \frac{π}{2} [

l n (t g (x)) = - 2 \sum_{k ⩾ 0} \frac{c o s (2 (2 k + 1) x)}{2 k + 1}

w e g e t

= - 2 \int_{0}^{\frac{π}{2}} x \sum_{k ⩾ 0} . - 2 \frac{c o s ((2 k + 1) x)}{2 k + 1} d x

= 4 \sum_{k ⩾ 0} \int_{0}^{\frac{π}{2}} x \frac{c o s ((2 k + 1) x)}{2 k + 1}, \int_{0}^{\frac{π}{2}} x \frac{c o s (2 k + 1) x}{2 k + 1} d x

= {[x \frac{s i n (2 k + 1) x}{{(2 k + 1)}^{2}}]}_{0}^{\frac{π}{2}} - \int_{0}^{\frac{π}{2}} \frac{s i n (2 k + 1) x}{{(2 k + 1)}^{2}} d x

= \frac{π}{2} . \frac{{(- 1)}^{k}}{{(2 k + 1)}^{2}} + {[\frac{c o s (2 k + 1) x}{{(2 k + 1)}^{3}}]}_{0}^{\frac{π}{2}} = \frac{π}{2} . \frac{{(- 1)}^{k}}{{(2 k + 1)}^{2}} - \frac{1}{{(2 k + 1)}^{3}}

4 \sum_{k ⩾ 0} \int_{0}^{\frac{π}{2}} x \frac{c o s ((2 k + 1) x)}{2 k + 1} = 4 \sum_{k ⩾ 0} (\frac{π}{2} . \frac{{(- 1)}^{k}}{{(2 k + 1)}^{2}} - \frac{1}{{(2 k + 1)}^{3}})

= 2 π Σ \frac{{(- 1)}^{k}}{{(2 k + 1)}^{2}} - 4 \sum_{k ⩾ 0} \frac{1}{{(2 k + 1)}^{3}}

G = \sum_{k ⩾ 0} \frac{{(- 1)}^{k}}{{(2 k + 1)}^{2}}, ζ (3) = Σ \frac{1}{k^{3}} = \underset{k = 0}{\sum^{\infty}} (\frac{1}{{(2 k + 2)}^{3}} + \frac{1}{{(2 k + 1)}^{3}})

= \frac{1}{8} \sum_{k ⩾ 0} \frac{1}{{(k + 1)}^{3}} + Σ \frac{1}{{(2 k + 1)}^{3}} \Rightarrow \sum_{k ⩾ 0} \frac{1}{{(2 k + 1)}^{3}} = \frac{7}{8} ζ (3)

w e g e t S o

2 π G - 4 . \frac{7}{8} ζ (3) = 2 π G - \frac{7}{2} ζ (3)

Commented by Tawa11 last updated on 06/Sep/21

great sir

Answered by maths mind last updated on 15/Sep/20

\int_{0}^{1} l n {[\frac{x + \sqrt{1 - x^{2}}}{x - \sqrt{1 - x^{2}}}]}^{2} \frac{x}{1 - x^{2}} d x, x = s i n (t)

= \int_{0}^{\frac{π}{2}} l n {(\frac{s i n (t) + c o s (t)}{s i n (t) - c o s (t)})}^{2} \frac{s i n (t) c o s (t)}{{c o s}^{2} (t)} d t

= \int_{0}^{\frac{π}{2}} l n {(\frac{1 + t g (t)}{- 1 + t g (t)})}^{2} t g (t) d t

= \int_{0}^{\frac{π}{4}} 2 l n (\frac{t g (t) + 1}{1 - t g (t)}) t g (t) d t + \int_{\frac{π}{4}}^{\frac{π}{2}} 2 l n (\frac{t g (t) + 1}{t g (t) - 1}) t g (t) d t

= 2 \int_{0}^{\frac{π}{4}} t g (t) l n (t g (\frac{π}{4} + t)) d t - 2 \int_{\frac{π}{4}}^{\frac{π}{2}} t g (t) l n (\frac{t g (t) - 1}{t g (t) + 1}) d t

= - 2 \int_{0}^{\frac{π}{4}} l n (t g (t)) t g (\frac{π}{4} - t) d t - 2 \int_{\frac{π}{4}}^{\frac{π}{2}} t g (t) l n (t g (t - \frac{π}{4})) {d t}_{t - \frac{π}{4} = r}

= - 2 \int_{0}^{\frac{π}{4}} l n (t g (t)) \frac{1 + t g (t)}{1 + t g (t)} d t - 2 \int_{0}^{\frac{π}{4}} t g (r + \frac{π}{4}) l n (t g (r)) d r

= - 2 \int_{0}^{\frac{π}{4}} l n (t g (t)) \frac{1 - t g (t)}{1 + t g (t)} d t - 2 \int_{0}^{\frac{π}{4}} l n (t g (t)) \frac{1 + t g (t)}{1 - t g (t)} d t

= - 2 \int_{0}^{\frac{π}{4}} l n (t g (t)) [\frac{1 - t g (t)}{1 + t g (t)} + \frac{1 + t g (t)}{1 - t g (t)}] d t

= - 2 \int_{0}^{\frac{π}{4}} [\frac{2 + 2 {t g}^{2} (t)}{1 - {t g}^{2} (t)}] l n (t g (t)) d t t g (t) = x \Rightarrow

= - 4 \int_{0}^{1} \frac{l n (x)}{1 - x^{2}} d x = - 4 \int_{0}^{1} l n (x) \sum_{k ⩾ 0} x^{2 k} d x

= - 4 \sum_{k ⩾ 0} \int_{0}^{1} x^{2 k} l n (x) = - 4 \sum_{k ⩾ 0} {{[\frac{x^{2 k + 1}}{2 k + 1} l n (x)]}_{0}^{1} - \frac{1}{2 k + 1} \int_{0}^{1} x^{2 k} d x

= 4 \sum_{k ⩾ 0} \frac{1}{{(2 k + 1)}^{2}} = 4 {ζ (2) - \frac{1}{4} ζ (2)} = 3 ζ (2) = 3 . \frac{π^{2}}{6} = \frac{π^{2}}{2}

Commented by Tawa11 last updated on 06/Sep/21

great sir

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