fint f(t)=∫_0 ^1 ((ln(1+tx^2 ))/x^2 )dx .


Question and Answers Forum

All Questions Topic List
Integration Questions
Previous in All Question Next in All Question
Previous in Integration Next in Integration
Question Number 55280 by Abdo msup. last updated on 20/Feb/19

$f i n t f (t) = \int_{0}^{1} \frac{l n (1 + {t x}^{2})}{x^{2}} d x .$
Commented by maxmathsup by imad last updated on 20/Feb/19
$we have f^′ (t) =∫_0 ^1 (∂/∂t)(((ln(1+tx^2 ))/x^2 ))dx =∫_0 ^1 (1/x^2 ) (x^2 /(1+tx^2 ))dx = ∫_0 ^1 (dx/(1+tx^2 )) so if t≥0 we get f^′ (t)=_((√t)x =u) ∫_0 ^(√t) (du/((√t)(1+u^2 ))) =(1/(√t))[arctan(u)]_0 ^(√t) =((arctan((√t)))/(√t)) ⇒f(t) =∫_0 ^t ((arctan((√u)))/(√u)) du +λ λ=f(0)=0 ⇒f(t) =∫_0 ^t ((arctan((√u)))/(√u)) du =_((√u)=x) ∫_0 ^(√t) ((arctanx)/x) (2x)dx =2 ∫_0 ^(√t) arctanx dx by parts f(t)=2{ [xarctanx]_0 ^(√t) −∫_0 ^(√t) (x/(1+x^2 ))dx} =2{(√t)arctan((√t)) −[(1/2)ln(1+x^2 )]_0 ^(√t) ⇒f(t)=2(√t)arctan((√t))−ln(1+t) . it t<0 we get f^′ (t) =∫_0 ^1 (dx/(1−(−t)x^2 )) =∫_0 ^1 (dx/((1−(√(−t))x)(1+(√(−t))x))) =(1/2)∫_0 ^1 ( (1/(1−(√(−t))x)) +(1/(1+(√(−t))x)))dx =(1/(2(√(−t)))) { ∫_0 ^1 ((√(−t))/(1−(√(−t))x)) +∫_0 ^1 ((√(−t))/(1+(√(−t))x))dx} =(1/(2(√(−t))))[ln∣1+(√(−t))x∣−ln∣1−(√(−t))x∣]_(x=0) ^(x=1) =(1/(2(√(−t)))) ln∣((1+(√(−t)))/(1−(√(−t))))∣ ⇒ f(t) =∫(1/(2(√(−t))))ln∣((1+(√(−t)))/(1−(√(−t))))∣dt +c =_((√(−t))=x) ∫ (1/(2x))ln∣((1+x)/(1−x))∣(−2x)dx +c = c−∫ ln∣1+x∣dx+∫ln∣1−x∣dx$
$w e h a v e f^{^{'}} (t) = \int_{0}^{1} \frac{\partial}{\partial t} (\frac{l n (1 + {t x}^{2})}{x^{2}}) d x = \int_{0}^{1} \frac{1}{x^{2}} \frac{x^{2}}{1 + {t x}^{2}} d x$ $= \int_{0}^{1} \frac{d x}{1 + {t x}^{2}} s o i f t ⩾ 0 w e g e t f^{^{'}} (t) =_{\sqrt{t} x = u} \int_{0}^{\sqrt{t}} \frac{d u}{\sqrt{t} (1 + u^{2})}$ $= \frac{1}{\sqrt{t}} {[a r c t a n (u)]}_{0}^{\sqrt{t}} = \frac{a r c t a n (\sqrt{t})}{\sqrt{t}} \Rightarrow f (t) = \int_{0}^{t} \frac{a r c t a n (\sqrt{u})}{\sqrt{u}} d u + λ$ $λ = f (0) = 0 \Rightarrow f (t) = \int_{0}^{t} \frac{a r c t a n (\sqrt{u})}{\sqrt{u}} d u =_{\sqrt{u} = x} \int_{0}^{\sqrt{t}} \frac{a r c t a n x}{x} (2 x) d x$ $= 2 \int_{0}^{\sqrt{t}} a r c t a n x d x b y p a r t s f (t) = 2 {{[x a r c t a n x]}_{0}^{\sqrt{t}} - \int_{0}^{\sqrt{t}} \frac{x}{1 + x^{2}} d x}$ $= 2 {\sqrt{t} a r c t a n (\sqrt{t}) - {[\frac{1}{2} l n (1 + x^{2})]}_{0}^{\sqrt{t}} \Rightarrow f (t) = 2 \sqrt{t} a r c t a n (\sqrt{t}) - l n (1 + t) .$ $i t t < 0 w e g e t f^{^{'}} (t) = \int_{0}^{1} \frac{d x}{1 - (- t) x^{2}} = \int_{0}^{1} \frac{d x}{(1 - \sqrt{- t} x) (1 + \sqrt{- t} x)}$ $= \frac{1}{2} \int_{0}^{1} (\frac{1}{1 - \sqrt{- t} x} + \frac{1}{1 + \sqrt{- t} x}) d x = \frac{1}{2 \sqrt{- t}} {\int_{0}^{1} \frac{\sqrt{- t}}{1 - \sqrt{- t} x} + \int_{0}^{1} \frac{\sqrt{- t}}{1 + \sqrt{- t} x} d x}$ $= \frac{1}{2 \sqrt{- t}} {[l n ∣ 1 + \sqrt{- t} x ∣ - l n ∣ 1 - \sqrt{- t} x ∣]}_{x = 0}^{x = 1}$ $= \frac{1}{2 \sqrt{- t}} l n ∣ \frac{1 + \sqrt{- t}}{1 - \sqrt{- t}} ∣ \Rightarrow f (t) = \int \frac{1}{2 \sqrt{- t}} l n ∣ \frac{1 + \sqrt{- t}}{1 - \sqrt{- t}} ∣ d t + c$ $=_{\sqrt{- t} = x} \int \frac{1}{2 x} l n ∣ \frac{1 + x}{1 - x} ∣ (- 2 x) d x + c = c - \int l n ∣ 1 + x ∣ d x + \int l n ∣ 1 - x ∣ d x$
	Answered by tm888 last updated on 20/Feb/19

Terms of Service
Privacy Policy
Contact: info@tinkutara.com

Question and Answers Forum