find ∫_0 ^1 (dx/(1+x^n )) interms of ψ (digamma)


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Question Number 174117 by Mathspace last updated on 25/Jul/22

$f i n d \int_{0}^{1} \frac{d x}{1 + x^{n}} i n t e r m s o f$ $ψ (d i g a m m a)$
	Answered by Mathspace last updated on 25/Jul/22
	$∫_0 ^1 (dx/(1+x^n ))=∫_0 ^1 Σ_(p=0) ^∞ (−1)^p x^(np) dx =Σ_(p=0) ^∞ (−1)^p ∫_0 ^1 x^(np) dx =Σ_(p=0) ^∞ (((−1)^p )/(np+1)) =Σ_(k=0) ^∞ (1/(2nk+1))−Σ_(k=0) ^∞ (1/((2k+1)n+1)) =Σ_(k=0) ^∞ ((1/(2nk+1))−(1/(2nk+n+1))) =nΣ_(k=0) ^∞ (1/((2nk+1)(2nk+n+1))) =(n/(4n^2 ))Σ_(k=0) ^∞ (1/((k+(1/(2n)))(k+((n+1)/(2n))))) =(1/(4n))(ψ(((n+1)/(2n)))−ψ((1/(2n))))×(1/(((n+1)/(2n))−(1/(2n)))) =(1/(2n)){ψ((1/(2n))+(1/2))−ψ((1/(2n)))}$
	$\int_{0}^{1} \frac{d x}{1 + x^{n}} = \int_{0}^{1} \sum_{p = 0}^{\infty} {(- 1)}^{p} x^{n p} d x$ $= \sum_{p = 0}^{\infty} {(- 1)}^{p} \int_{0}^{1} x^{n p} d x$ $= \sum_{p = 0}^{\infty} \frac{{(- 1)}^{p}}{n p + 1}$ $= \sum_{k = 0}^{\infty} \frac{1}{2 n k + 1} - \sum_{k = 0}^{\infty} \frac{1}{(2 k + 1) n + 1}$ $= \sum_{k = 0}^{\infty} (\frac{1}{2 n k + 1} - \frac{1}{2 n k + n + 1})$ $= n \sum_{k = 0}^{\infty} \frac{1}{(2 n k + 1) (2 n k + n + 1)}$ $= \frac{n}{4 n^{2}} \sum_{k = 0}^{\infty} \frac{1}{(k + \frac{1}{2 n}) (k + \frac{n + 1}{2 n})}$ $= \frac{1}{4 n} (ψ (\frac{n + 1}{2 n}) - ψ (\frac{1}{2 n})) \times \frac{1}{\frac{n + 1}{2 n} - \frac{1}{2 n}}$ $= \frac{1}{2 n} {ψ (\frac{1}{2 n} + \frac{1}{2}) - ψ (\frac{1}{2 n})}$
	Commented by aleks041103 last updated on 25/Jul/22

	$Y e s, t h i s i s c o r r e c t! I m a d e a n e x t r e m e l y$ $b i g m i s t a k e o f d o i n g w h a t e v e r I w a n t w i t h$ $n o n c o n v e r g i n g i n t e g r a l s . . .$
	Commented by Mathspace last updated on 25/Jul/22

	$y o u r m e t h o d i s c o r r e c t b y y o u h s v e c o m m i t e d$ $a e r r o r o f c a l c u l u s n e v e r m i n d s i r$
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