let-f-a-0-ln-x-x-2-a-with-a-gt-0-1-calculate-f-a-intermsof-a-2-find-the-values-of-0-ln-x-x-2-1-dx-and-0-ln-x-x-2-2-dx-3-let-g-a-0-ln-x-

Question Number 55454 by maxmathsup by imad last updated on 24/Feb/19

l e t f (a) = \int_{0}^{\infty} \frac{l n (x)}{x^{2} + a} w i t h a > 0

1) c a l c u l a t e f (a) i n t e r m s o f a

2) f i n d t h e v a l u e s o f \int_{0}^{\infty} \frac{l n (x)}{x^{2} + 1} d x a n d \int_{0}^{\infty} \frac{l n (x)}{x^{2} + 2} d x

3) l e t g (a) = \int_{0}^{\infty} \frac{l n (x)}{{(x^{2} + a)}^{n}} d x . c a l c u l a t e g (a) i n t e r m s o f a

4) f i n d v a l u e s > o f \int_{0}^{\infty} \frac{l n (x)}{{(x^{2} + 3)}^{2}} d x

5) f i n d n a t u r e o f t h e s e r i e Σ f (n) a n d Σ g (n)

Commented by maxmathsup by imad last updated on 24/Feb/19

g (a) = \int_{0}^{\infty} \frac{l n (x)}{{(x^{2} + a)}^{2}} d x .

Commented by maxmathsup by imad last updated on 26/Feb/19

1) c h a n g e m e n t x = \sqrt{a} t g i v e f (a) = \int_{0}^{\infty} \frac{l n (\sqrt{a} t)}{a (t^{2} + 1)} \sqrt{a} d t = \frac{1}{\sqrt{a}} \int_{0}^{\infty} \frac{l n (\sqrt{a}) + l n (t)}{1 + t^{2}} d t

= \frac{l n (\sqrt{a)}}{\sqrt{a}} \int_{0}^{\infty} \frac{d t}{1 + t^{2}} + \frac{1}{\sqrt{a}} \int_{0}^{\infty} \frac{l n (t)}{1 + t^{2}} d t b u t c h a n g . t = \frac{1}{u} d r i v e t o \int_{0}^{\infty} \frac{l n (t)}{1 + t^{2}} d t = 0 \Rightarrow

f (a) = \frac{π}{2} \frac{l n (\sqrt{a})}{\sqrt{a}} \Rightarrow ★ f (a) = \frac{π}{4} \frac{l n (a)}{\sqrt{a}} ★

2) w e h a v e \int_{0}^{\infty} \frac{l n (x)}{x^{2} + 1} d x = f (1) = 0

\int_{0}^{\infty} \frac{l n (x)}{x^{2} + 2} d x = f (2) = \frac{π l n (2)}{4 \sqrt{2}}

3) w e h a v e f^{^{'}} (a) = - \int_{0}^{\infty} \frac{l n (x)}{{(x^{2} + a)}^{2}} d x = - g (a) \Rightarrow g (a) = - f^{^{'}} (a)

b u t f^{^{'}} (a) = \frac{π}{4} \frac{\frac{\sqrt{a}}{a} - l n (a) \frac{1}{2 \sqrt{a}}}{a} = \frac{π}{4} \frac{\frac{2 a - a l n (a)}{2 a \sqrt{a}}}{a} = \frac{π}{4} \frac{2 - l n a}{2 a \sqrt{a}} \Rightarrow

★ g (a) = \frac{π (l n (a) - 2)}{8 a \sqrt{a}} ★

4) \int_{0}^{\infty} \frac{l n (x)}{{(x^{2} + 3)}^{2}} d x = g (3) = \frac{π (l n 3 - 2)}{24 \sqrt{3}}

Commented by maxmathsup by imad last updated on 26/Feb/19

5) w e h a v e \sum_{n ⩾ 1} f (n) = \frac{π}{4} \sum_{n ⩾ 1} \frac{l n (n)}{\sqrt{n}}

l e t φ (t) = \frac{l n (t)}{\sqrt{t}} \Rightarrow φ^{^{'}} (t) = \frac{\frac{\sqrt{t}}{t} - l n (t) \frac{1}{2 \sqrt{t}}}{t} = \frac{\frac{1}{\sqrt{t}} - \frac{l n (t)}{2 \sqrt{t}}}{t} = \frac{2 - l n (t)}{2 t \sqrt{t}}

f o r t ⩾ e^{2} \Rightarrow l n (t) ⩾ 2 \Rightarrow φ^{^{'}} (t) ⩽ 0 \Rightarrow φ i s d e c r e a s i n g s o Σ f (n) a n d

\int_{e^{2}}^{+ \infty} \frac{l n (x)}{\sqrt{x}} d x h a v e t h e s a m e n a t u r e

\int_{e^{2}}^{+ \infty} \frac{l n (x)}{\sqrt{x}} d x =_{x = t^{2}} \int_{e}^{+ \infty} \frac{2 l n (t)}{t} (2 t) d t = 4 \int_{e}^{+ \infty} l n (t) d t = + \infty \Rightarrow

Σ f (n) i s d i v e r g e n t .