advanced-calculud-compute-i-

Question Number 127315 by mnjuly1970 last updated on 28/Dec/20

\dots a d v a n c e d c a l c u l u d \dots

c o m p u t e :::

ψ (i) = ? ?

Commented by Dwaipayan Shikari last updated on 28/Dec/20

ψ (i) = - γ + \int_{0}^{1} \frac{1 - x^{i - 1}}{1 - x} d x

= - γ + \int_{0}^{1} \frac{x - x^{i}}{x} + \frac{x - x^{i}}{1 - x} d x

= - γ + 1 + i + \int_{0}^{1} \frac{x - x^{i}}{1 - x} d x

I (a) = \int_{0}^{1} \frac{x - x^{i}}{1 - x} d x = \int_{0}^{1} (\frac{1 - x - {(1 - x)}^{i}}{x}) e^{- a x} d x

I^{'} (a) = \int_{0}^{1} {(1 - x)}^{i} e^{- a x} d x = \int_{0}^{1} x^{i} e^{- a (1 - x)} d x = e^{- a} \int_{0}^{1} x^{i} e^{x} d x

= e^{- a} \underset{n = 0}{\sum^{\infty}} \frac{1}{n! (n + i)}

I (a) = - e^{- a} \underset{n = 1}{\sum^{\infty}} \frac{1}{n! (n + i)} + C \Rightarrow I (0) = - \underset{n = 0}{\sum^{\infty}} \frac{1}{n! (n + i)}

ψ (i) = 1 - γ + i - \underset{n = 0}{\sum^{\infty}} \frac{1}{n! (n + i)} \dots .

Commented by dsardor2004 last updated on 28/Dec/20

?

Answered by Olaf last updated on 28/Dec/20

U s i n g s e r i e s :

Re (ψ (i)) = - γ - \underset{n = 0}{\sum^{\infty}} \frac{n - 1}{n^{3} + n^{2} + n + 1}

Re (ψ (i)) = - γ + 1 + \underset{n = 2}{\sum^{\infty}} \frac{1}{n^{4} - 1}

Im (ψ (i)) = \underset{n = 0}{\sum^{\infty}} \frac{1}{n^{2} + 1} = \frac{1}{2} + \frac{π}{2} \cot π

ψ (i) = - γ + 1 + \underset{n = 2}{\sum^{\infty}} \frac{1}{n^{4} - 1} + i \frac{1 + π \cot π}{2}

Commented by mnjuly1970 last updated on 28/Dec/20

t h a n k y o u \dots

Answered by mindispower last updated on 28/Dec/20

Ψ (i) = - γ + \sum_{n ⩾ 1} (\frac{1}{n + 1} - \frac{1}{n + i})

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

= - γ + \sum_{n ⩾ 0} \frac{- n + 1}{n^{3} + n^{2} + n + 1} + i \sum_{n ⩾ 0} \frac{1}{n^{2} + 1}

R e Ψ (i) = - γ - \sum_{n ⩾ 0} \frac{n - 1}{n^{3} + n^{2} + n + 1}

I m (Ψ (i)) = \sum_{n ⩾ 0} \frac{1}{n^{2} + 1} = \frac{1}{2} (\underset{- \infty}{\sum^{\infty}} \frac{1}{n^{2} + 1} + 1)

= \frac{1}{2} (1 - 2 R e π c o t h (i π) . \frac{1}{1 - i})

= \frac{1}{2} (1 - R e (π i c o t h (π) . (1 + i))

= \frac{1}{2} (1 + π c o t t h (π))

I m (Ψ (i)) = \frac{1}{2} + \frac{π}{2} c o t h (π)

R e (Ψ (i)) = - γ - \sum_{n ⩾ 1} \frac{n - 1}{n^{3} + n^{2} + n + 1} . . n o c l o s e f o r m

i n e l e m e n t r y f u n c t i o n

Commented by mnjuly1970 last updated on 28/Dec/20

t h a n k y o u s o m u c h s i r p o w e r . .

Commented by mindispower last updated on 29/Dec/20

a l w a y s p l e a s u r s i r