nice-calculus-Evaluate-0-1-sin-ln-x-ln-x-ln-2-x-dx-Ans-ln-2-pi-4-1-

Question Number 126465 by mnjuly1970 last updated on 20/Dec/20

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

E v a l u a t e \dots

ϕ = \int_{0}^{1} \frac{s i n (l n (x)) - l n (x)}{{l n}^{2} (x)} d x

A n s :: l n (\sqrt{2}) + \frac{π}{4} - 1 \dots

Commented by talminator2856791 last updated on 22/Dec/20

do you have more difficult question than this

Answered by mindispower last updated on 23/Dec/20

- l n (x) = t

= \int_{0}^{\infty} \frac{t - s i n (t)}{t^{2}} e^{- t}

s i n (t) - t = \sum_{k ⩾ 1} \frac{{(- 1)}^{k} t^{2 k + 1}}{(2 k + 1)!}

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

= \sum_{k ⩾ 1} \frac{{(- 1)}^{k}}{(2 k + 1)!} Γ (2 k)

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

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

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

= - \frac{1}{2} \int_{0}^{1} \frac{1}{1 + x} d x - \int_{0}^{1} \frac{- x^{2}}{1 + x^{2}} d x

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

= 1 + \frac{π}{4} - l n (\sqrt{2})

Commented by mnjuly1970 last updated on 24/Dec/20

t h a n k s a l o t . .