Nice-Calculus-if-n-0-1-x-n-1-x-dx-then-n-1-1-n-1-n-n-

Question Number 144231 by mnjuly1970 last updated on 23/Jun/21

\dots . . N i c e \dots C a l c u l u s

i f ::

φ (n) := \int_{0}^{1} \frac{x^{n}}{1 + x} d x

t h e n :: \underset{n = 1}{\sum^{\infty}} \frac{{(- 1)}^{n - 1} φ (n)}{n} = ?

\dots \dots . .

Answered by mathmax by abdo last updated on 23/Jun/21

φ (n) = \int_{0}^{1} \frac{x^{n}}{1 + x} dx \Rightarrow let S = \sum_{n = 1}^{\infty} \frac{{(- 1)}^{n - 1}}{n} φ (n)

= \sum_{n = 1}^{\infty} \frac{{(- 1)}^{n - 1}}{n} \int_{0}^{1} \frac{x^{n}}{1 + x} dx = \int_{0}^{1} (\sum_{n = 1}^{\infty} \frac{{(- 1)}^{n - 1}}{n} \frac{x^{n}}{1 + x}) dx

= \int_{0}^{1} \frac{w (x)}{1 + x} dx with w (x) = \sum_{n = 1}^{\infty} \frac{{(- 1)}^{n - 1}}{n} x^{n} \Rightarrow

w^{^{'}} (x) = \sum_{n = 1}^{\infty} {(- 1)}^{n - 1} x^{n - 1} = \sum_{n = 1}^{\infty} {(- x)}^{n - 1}

= \sum_{n = 0}^{\infty} {(- x)}^{n} = \frac{1}{1 + x} \Rightarrow w (x) = \log (1 + x) + c

w (0) = 0 = c \Rightarrow w (x) = \log (1 + x) \Rightarrow

S = \int_{0}^{1} \frac{\log (1 + x)}{1 + x} dx = {[\log^{2} (1 + x)]}_{0}^{1} - \int_{0}^{1} \frac{\log (1 + x)}{1 + x} dx \Rightarrow

2 \int_{0}^{1} \frac{\log (1 + x)}{1 + x} dx = \log^{2} (2) \Rightarrow S = \frac{\log^{2} (2)}{2}

Commented by mnjuly1970 last updated on 23/Jun/21

t h a n k s a l o t m r m a x \dots

Commented by mathmax by abdo last updated on 24/Jun/21

you are welcome sir .