calculate-0-ln-1-x-1-t-2-1-t-2-dt-with-x-gt-0-2-find-the-value-of-0-ln-2-t-2-1-t-2-dt-and-0-ln-3-2t-2-1-t-2-dt-

Question Number 116557 by Bird last updated on 04/Oct/20

c a l c u l a t e \int_{0}^{\infty} \frac{l n (1 + x (1 + t^{2})}{1 + t^{2}} d t

w i t h x > 0

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

a n d \int_{0}^{\infty} \frac{l n (3 + 2 t^{2})}{1 + t^{2}} d t

Answered by mindispower last updated on 05/Oct/20

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

f^{'} (x) = \int_{0}^{\infty} \frac{d t}{1 + x (1 + t^{2})}

= \int_{0}^{\infty} \frac{d t}{(1 + x) (1 + {(\frac{t}{\sqrt{1 + x}})}^{2})}

= \frac{1}{\sqrt{1 + x}} {[\tan^{- 1} (\frac{t}{\sqrt{1 + x}})]}_{0}^{\infty}

= \frac{π}{2 \sqrt{1 + x}}

f (0) = 0

f (x) = \int_{0}^{x} \frac{π}{2 \sqrt{1 + x}} = π \sqrt{1 + x}

\int_{0}^{\infty} \frac{l n (2 + t^{2})}{1 + t^{2}} d t = f (1) = π \sqrt{2}

\int_{0}^{\infty} \frac{l n (3 + 2 t^{2})}{1 + t^{2}} d t = f (2) = π \sqrt{3}

Commented by mathmax by abdo last updated on 05/Oct/20

thanks sir

Commented by mathmax by abdo last updated on 05/Oct/20

(1 + x) (1 + {(\frac{t}{\sqrt{1 + x}})}^{2}) = (1 + x) {1 + \frac{t^{2}}{1 + x}} = 1 + x + t^{2} \neq 1 + x (1 + t^{2})!

Answered by mathmax by abdo last updated on 05/Oct/20

1) let f (x) = \int_{0}^{\infty} \frac{\ln (1 + x (1 + t^{2}))}{1 + t^{2}} dt we have f^{^{'}} (x) = \int_{0}^{\infty} \frac{dt}{1 + x (1 + t^{2})}

= \int_{0}^{\infty} \frac{dt}{{xt}^{2} + x + 1} = \frac{1}{x} \int_{0}^{\infty} \frac{dt}{t^{2} + \frac{x + 1}{x}} =_{t = \sqrt{\frac{x + 1}{x}} u} \frac{1}{x} . \frac{x}{x + 1} \int_{0}^{\infty} \frac{1}{u^{2} + 1} \frac{\sqrt{x + 1}}{\sqrt{x}} du

= \frac{1}{\sqrt{x} \sqrt{x + 1}} \Rightarrow f (x) = \int \frac{dx}{\sqrt{x} \sqrt{x + 1}} + c

=_{\sqrt{x} = z} \int \frac{2 zdz}{z \sqrt{z^{2} + 1}} + c = 2 \int \frac{dz}{\sqrt{z^{2} + 1}} + c

= 2 \ln (z + \sqrt{1 + z^{2}}) + c = 2 \ln (\sqrt{x} + \sqrt{1 + x}) + c

f (0) = 0 = 2 \ln (1) + c = c \Rightarrow f (x) = 2 \ln (\sqrt{x} + \sqrt{1 + x})

2) \int_{0}^{\infty} \frac{\ln (2 + t^{2})}{1 + t^{2}} dt = f (1) = 2 \ln (1 + \sqrt{2})

\int_{0}^{\infty} \frac{\ln (3 + {2 t}^{2})}{1 + t^{2}} dt = f (2) = 2 \ln (\sqrt{2} + \sqrt{3})