challanging-integral-prove-that-0-cos-x-1-1-x-2-dx-x-

Question Number 127446 by mnjuly1970 last updated on 29/Dec/20

\dots c h a l l a n g i n g i n t e g r a l \dots

p r o v e t h a t ::

Ω = \int_{0}^{\infty} (c o s (x) - \frac{1}{1 + x^{2}}) \frac{d x}{x} = - γ

Answered by mindispower last updated on 31/Dec/20

C i (x) = - \int_{x}^{\infty} \frac{c o s (t)}{t} d t = γ + l n (x) - \int_{0}^{x} \frac{1 - c o s (t)}{t} d t

Ω = \lim_{t \to 0} \int_{t}^{\infty} (c o s (x) - \frac{1}{1 + x^{2}}) . \frac{d x}{x}

= \lim_{t \to 0} (- γ - l n (t) + \int_{0}^{t} \frac{1 - c o s (x)}{x} d x - \int_{t}^{\infty} \frac{d x}{x (1 + x^{2})})

= \lim_{t \to 0} (- γ - l n (t) + \int_{0}^{t} \frac{1 - c o s (x)}{x} d x - B)

B = \int_{t}^{\infty} \frac{1}{x (1 + x^{2})} = \frac{1 + x^{2} - x . x}{x (1 + x^{2})} = \int_{t}^{\infty} \frac{1}{x} - \frac{x}{1 + x^{2}} d x

= {[l n (x) - \frac{l n (1 + x^{2})}{2}]}_{t}^{\infty} = - l n (t) + \frac{l n (1 + t^{2})}{2}

Ω = \lim_{t \to 0} (- γ - l n (t) + \int_{0}^{t} \frac{1 - c o s (x)}{x} d x + l n (t) - \frac{l n (1 + t^{2})}{2})

Ω = \lim_{t \to 0} (- γ + \int_{0}^{t} \frac{1 - c o s (x)}{x} d x)

\lim_{t \to 0} \int_{0}^{t} \frac{1 - c o s (x)}{x} d x = 0

\frac{1 - c o s (x)}{x} = g (x)

0 ⩽ \frac{1 - c o s (x)}{x} \underset{x \in [0, t]}{} = \sum_{k ⩾ 1} \frac{{(- 1)}^{k + 1}}{(2 k!)} x^{2 k - 1} ⩽ \frac{x}{2}

\Rightarrow 0 ⩽ \int_{0}^{t} \frac{1 - c o s (x)}{x} d x ⩽ \frac{t^{2}}{4} \to 0

Ω = \lim_{t \to 0} (- γ + \int_{0}^{t} \frac{1 - c o s (x)}{x} d x) = - γ + 0 = - γ

Ω = - γ

Commented by mnjuly1970 last updated on 31/Dec/20

v e r y n i c e . t h a n k s a l o t . .

Commented by mindispower last updated on 31/Dec/20

p l e a s u r s i r h a p p y n e w y e a r s a l l g o o d t h i n g s f o r

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