nice-calculus-find-a-series-representation-for-the-following-integral-0-1-cos-h-xln-x-dx-

Question Number 123898 by mnjuly1970 last updated on 29/Nov/20

\dots n i c e c a l c u l u s \dots

f i n d a s e r i e s r e p r e s e n t a t i o n

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

ϕ = \int_{0}^{1} c o s (h (x l n (x)) d x

Answered by mindispower last updated on 29/Nov/20

\int_{0}^{1} {l n}^{a} (x) d x = \int_{0}^{\infty} {(- t)}^{a} e^{- t} d t \dots 1

= {(- 1)}^{a} Γ (a + 1) \dots . . n i c e

c o s (x) = \sum_{n ⩾ 0} \frac{{(- 1)}^{n} x^{2 n}}{(2 n)!}

\emptyset = \int_{0}^{1} \sum_{n ⩾ 0} \frac{{(- 1)}^{n}}{2 n!} . {(h l n (x))}^{2 n} d x

= \sum_{n ⩾ 0} \frac{{(- 1)}^{n}}{(2 n)!} h^{2 n} \int_{0}^{1} {l n}^{2 n} (x) d x \dots . a p p i e 1 a = 2 n

= \sum_{n ⩾ 0} \frac{{(- 1)}^{n}}{(2 n)!} h^{2 n} . \frac{{(- 1)}^{2 n} Γ (2 n + 1)}{}

= \sum_{n ⩾ 0} {(- h^{2})}^{n} = \frac{1}{1 + h^{2}}

Commented by Mammadli last updated on 29/Nov/20

Commented by Dwaipayan Shikari last updated on 29/Nov/20

\lim_{x \to 1} \frac{a (a^{x - 1} - 1)}{x - 1} = a . l o g (a)

Commented by mnjuly1970 last updated on 29/Nov/20

t h a n k s f o r y o u r w o r k

s i r p o w e r . b u t

h : h y p b o l i c

Answered by Dwaipayan Shikari last updated on 29/Nov/20

\int_{0}^{1} c o s (h (l o g (x)) d x

= \int_{0}^{1} \frac{e^{l o g x} + e^{- l o g x}}{2} d x

= \frac{1}{2} \int_{0}^{1} \underset{n = 0}{\sum^{\infty}} \frac{{(l o g x)}^{n}}{n!} + \frac{1}{2} \int_{0}^{1} \underset{n = 0}{\sum^{\infty}} \frac{{(- 1)}^{n} {(l o g x)}^{n}}{n!}

= \frac{1}{2} (\underset{n = 0}{\sum^{\infty}} \frac{1}{n!} \int_{0}^{1} {(l o g x)}^{n} + \underset{n = 0}{\sum^{\infty}} \frac{{(- 1)}^{n}}{n!} \int_{0}^{1} {(l o g x)}^{n})

= \frac{1}{2} \underset{n = 0}{\sum^{\infty}} \frac{1}{n!} \int_{- \infty}^{0} e^{t} t^{n} + \frac{1}{2} \underset{n = 0}{\sum^{\infty}} \frac{{(- 1)}^{n}}{n!} \int_{- \infty}^{0} e^{t} t^{n} d t t = - Φ

= \frac{1}{2} \underset{n = 0}{\sum^{\infty}} \frac{{(- 1)}^{n}}{n!} \int_{0}^{\infty} e^{- Φ} Φ^{n} d Φ + \frac{1}{2} \underset{n = 0}{\sum^{\infty}} \frac{1}{n!} \int_{0}^{\infty} e^{- Φ} Φ^{n} d Φ

= \underset{n = 0}{\sum^{\infty}} {(- 1)}^{n} \frac{Γ (n + 1)}{2 n!} + \underset{n = 0}{\sum^{\infty}} \frac{Γ (n + 1)}{2 n!}

Commented by mnjuly1970 last updated on 29/Nov/20

t h a n k y o u s o m u c h s i r

p a y a n .

q u e s t i o n w a s c o r r e c t e d :

i n t e g r a n d : c o s (h (x l n (x)))

Answered by Dwaipayan Shikari last updated on 29/Nov/20

\int_{0}^{1} c o s (h (x l o g x)) d x

= \int_{0}^{1} \frac{e^{x l o g x} + e^{- x l o g x}}{2} d x

= \frac{1}{2} \underset{n ⩾ 0}{\sum^{\infty}} \int_{0}^{1} \frac{{(x l o g x)}^{n}}{n!} - \frac{1}{2} \sum_{n ⩾ 0} {(- 1)}^{n} \int_{0}^{1} \frac{{(x l o g x)}^{n}}{n!} (t (n + 1) = - Φ)

= \frac{1}{2} \sum_{n ⩾ 0} \frac{1}{n!} \int_{- \infty}^{0} e^{(n + 1) t} t^{n} d t - \frac{1}{2} \sum_{n ⩾ 0} {(- 1)}^{n} \frac{1}{n!} \int_{- \infty}^{0} e^{t (n + 1)} t^{n} d t

= \frac{1}{2} \sum_{n ⩾ 0} \frac{1}{n!} \frac{(- 1)}{{(n + 1)}^{n + 1}} \int_{0}^{\infty} e^{- Φ} Φ^{n} d Φ - \frac{1}{2} \sum_{n ⩾ 0} \frac{{(- 1)}^{2 n}}{n! {(n + 1)}^{n + 1}} \int_{0}^{\infty} e^{- Φ} Φ^{n} d Φ

= \frac{1}{2} \sum_{n ⩾ 0} \frac{{(- 1)}^{n}}{{(n + 1)}^{n + 1}} - \frac{1}{2} \sum_{n ⩾ 0} \frac{1}{{(n + 1)}^{(n + 1)}}

= \frac{1}{2} (\frac{1}{1^{1}} - \frac{1}{2^{2}} + \frac{1}{3^{3}} - \frac{1}{4^{4}} + \dots - \frac{1}{1^{1}} - \frac{1}{2^{2}} - \frac{1}{3^{3}} - \frac{1}{4^{4}} + - . .)

= - (\frac{1}{2^{2}} + \frac{1}{4^{4}} + \frac{1}{6^{6}} + \frac{1}{8^{8}} + \dots .) = - \sum_{n ⩾ 1} \frac{1}{{(2 n)}^{2 n}}

Commented by mnjuly1970 last updated on 29/Nov/20

p e a c e b e u p o n y o u

s i r p a y a n

t i r e l e s s, p o w e r f u l

a n d u n a s s u m i n g .

g r a t e f u l a n d g o d k e e p y o u . .

Commented by Dwaipayan Shikari last updated on 29/Nov/20

W i t h p l e a s u r e s i r! y o u a r e r e a l l y g r e a t m a n