let-f-x-0-x-dt-1-t-4-1-find-a-explicit-form-of-f-x-2-calculate-0-dt-1-t-4-

Question Number 43623 by math khazana by abdo last updated on 12/Sep/18

l e t f (x) = \int_{0}^{x} \frac{d t}{1 + t^{4}}

1) f i n d a e x p l i c i t f o r m o f f (x)

2) c a l c u l a t e \int_{0}^{\infty} \frac{d t}{1 + t^{4}}

Commented by maxmathsup by imad last updated on 13/Sep/18

l e t d e c o m p o s e F (t) = \frac{1}{t^{4} + 1} w e h a v e F (t) = \frac{1}{{(t^{2} + 1)}^{2} - 2 t^{2}}

= \frac{1}{(t^{2} + 1 - \sqrt{2} t) (t^{2} + 1 + \sqrt{2} t)} = \frac{a t + b}{(t^{2} - \sqrt{2} t + 1)} + \frac{c t + d}{t^{2} + \sqrt{2} t + 1}

F (- t) = F (t) \Rightarrow \frac{- a t + b}{t^{2} + \sqrt{2} t + 1} + \frac{- c t + d}{t^{2} - \sqrt{2} t + 1} = F (t) \Rightarrow c = - a a n d b = d \Rightarrow

F (x) = \frac{a t + b}{t^{2} - \sqrt{2} t + 1} + \frac{- a t + b}{t^{2} + \sqrt{2} t + 1}

F (0) = 1 = 2 b \Rightarrow b = 0 \Rightarrow F (t) = \frac{a t}{t^{2} - \sqrt{2} t + 1} - \frac{a t}{t^{2} + \sqrt{2} t + 1}

F (1) = \frac{1}{2} = \frac{a}{2 - \sqrt{2}} - \frac{a}{2 + \sqrt{2}} = (\frac{2 + \sqrt{2} - 2 + \sqrt{2}}{2}) a = \sqrt{2} a \Rightarrow a = \frac{1}{2 \sqrt{2}} \Rightarrow

F (x) = \frac{1}{2 \sqrt{2}} {\frac{t}{t^{2} - \sqrt{2} t + 1} - \frac{t}{t^{2} + \sqrt{2} t + 1}} \Rightarrow

\int \frac{d t}{1 + t^{4}} = \frac{1}{4 \sqrt{2}} {\int \frac{2 t - \sqrt{2} + \sqrt{2}}{t^{2} - \sqrt{2} t + 1} d t - \int \frac{2 t + \sqrt{2} - \sqrt{2}}{t^{2} + \sqrt{2} t + 1} d t}

\Rightarrow 4 \sqrt{2} \int \frac{d t}{1 + t^{4}} = l n ∣ t^{2} - \sqrt{2} t + 1 ∣ - l n ∣ t^{2} + \sqrt{2} t + 1 ∣ + \sqrt{2} \int \frac{d t}{t^{2} - \sqrt{2} t + 1} - \sqrt{2} \int \frac{d t}{t^{2} + \sqrt{2} t + 1}

b u t \int \frac{d t}{t^{2} - \sqrt{2} t + 1} = \int \frac{d t}{t^{2} - \frac{2}{\sqrt{2}} t + \frac{1}{2} + \frac{1}{2}} = \int \frac{d t}{{(t - \frac{1}{\sqrt{2}})}^{2} + \frac{1}{2}}

=_{t - \frac{1}{\sqrt{2}} = \frac{u}{\sqrt{2}}} \int \frac{2}{1 + u^{2}} \frac{d u}{\sqrt{2}} = \sqrt{2} a r c t a n (u) = \sqrt{2} a r c t a n (t \sqrt{2} - 1) a l s o

\int \frac{d t}{t^{2} + \sqrt{2} t + 1} = \sqrt{2} a r c t a n (t \sqrt{2} + 1) \Rightarrow

4 \sqrt{2} \int \frac{d t}{1 + t^{4}} = l n ∣ \frac{t^{2} - \sqrt{2} t + 1}{t^{2} + \sqrt{2} t + 1} ∣ + 2 a r c t a n (t \sqrt{2} - 1) - 2 a r c t a n (t \sqrt{2} + 1) + c \Rightarrow

\int \frac{d t}{1 + t^{4}} = \frac{1}{4 \sqrt{2}} l n ∣ \frac{t^{2} - \sqrt{2} t + 1}{t^{2} + \sqrt{2} t + 1} ∣ + \frac{1}{2 \sqrt{2}} {a r c t a n (t \sqrt{2} - 1) - a r c t a n (t \sqrt{2} + 1)} + c \Rightarrow

f (x) = \int_{0}^{x} \frac{d t}{1 + t^{4}} = [\frac{1}{4 \sqrt{2}} l n ∣ \frac{t^{2} - \sqrt{2} t + 1}{t^{2} + \sqrt{2} t + 1} ∣_{0}^{x} + \frac{1}{2 \sqrt{2}} {[a r c t a n (t \sqrt{2} - 1) - a r c t a n (t \sqrt{2} + 1)]}_{0}^{x}

f (x) = \frac{1}{4 \sqrt{2}} l n ∣ \frac{x^{2} - \sqrt{2} x + 1}{x^{2} + \sqrt{2} x + 1} ∣ + \frac{1}{2 \sqrt{2}} {a r c t a n (x \sqrt{2} - 1) - a r c t a n (x \sqrt{2} + 1) + \frac{π}{2}}

Commented by maxmathsup by imad last updated on 13/Sep/18

e r r o r a t l i n e 9

4 \sqrt{2} \int \frac{d t}{1 + t^{4}} = l n ∣ t^{2} - \sqrt{2} t + 1 ∣ - l n ∣ t^{2} + \sqrt{2} t + 1 ∣ + \sqrt{2} \int \frac{d t}{t^{2} - \sqrt{2} t + 1} + \sqrt{2} \int \frac{d t}{t^{2} + \sqrt{2} t + 1}

\Rightarrow

f (x) = \frac{1}{4 \sqrt{2}} l n ∣ \frac{x^{2} - \sqrt{2} x + 1}{x^{2} + \sqrt{2} x + 1} ∣ + \frac{1}{2 \sqrt{2}} {a r c t a n (x \sqrt{2} - 1) + a r c t a n (x \sqrt{2} + 1)} .

Commented by maxmathsup by imad last updated on 13/Sep/18

2) w e h a v e \int_{0}^{\infty} \frac{d t}{1 + t^{4}} = {l i m}_{x \to + \infty} f (x) = \frac{1}{2 \sqrt{2}} {\frac{π}{2} + \frac{π}{2}} = \frac{π}{2 \sqrt{2}} .

Answered by tanmay.chaudhury50@gmail.com last updated on 13/Sep/18

\int \frac{d t}{1 + t^{4}}

\int \frac{\frac{1}{t^{2}}}{t^{2} + \frac{1}{t^{2}}} d t

\frac{1}{2} \int \frac{(1 + \frac{1}{t^{2}}) - (1 - \frac{1}{t^{2}})}{t^{2} + \frac{1}{t^{2}}} d t

\frac{1}{2} \int \frac{d (t - \frac{1}{t})}{{(t - \frac{1}{t})}^{2} + 2} - \frac{1}{2} \int \frac{d (t + \frac{1}{t})}{{(t + \frac{1}{t})}^{2} - 2}

I_{1} = \frac{1}{2} \int \frac{d (t - \frac{1}{t})}{{(t - \frac{1}{t})}^{2} + 2}

= \frac{1}{2} \times \frac{1}{\sqrt{2}} {t a n}^{- 1} (\frac{t - \frac{1}{t}}{\sqrt{2}}) + c

I_{2} = \frac{1}{2} \int \frac{d (t + \frac{1}{t})}{{(t + \frac{1}{t})}^{2} - {(\sqrt{2})}^{2}}

f o r m u l a \int \frac{d y}{y^{2} - a^{2}} = \frac{1}{2 a} \int \frac{(y + a) - (y - a)}{(y + a) y - a)} d y

\frac{1}{2 a} [\int \frac{d y}{y - a} - \int \frac{d y}{y + a}]

\frac{1}{2 a} l n ∣ \frac{y - a}{y + a} ∣ + c_{2}

I_{2} = \frac{1}{2} \times \frac{1}{2 \sqrt{2}} l n ∣ \frac{t + \frac{1}{t} - \sqrt{2}}{t + \frac{1}{t} + \sqrt{2}} ∣

s o a n s i s I_{1} - I_{2}

\frac{1}{2 \sqrt{2}} {t a n}^{- 1} (\frac{t - \frac{1}{t}}{\sqrt{2}}) - \frac{1}{4 \sqrt{2}} l n ∣ \frac{t + \frac{1}{t} - \sqrt{2}}{t + \frac{1}{t} + \sqrt{2}} ∣ + c

s o

\int_{0}^{x} \frac{d t}{1 + t^{4}}

[\frac{1}{2 \sqrt{2}} {t a n}^{- 1} (\frac{x - \frac{1}{x}}{\sqrt{2}}) - \frac{1}{4 \sqrt{2}} l n ∣ \frac{x + \frac{1}{x} - \sqrt{2}}{x + \frac{1}{x} + \sqrt{2}} ∣] -

[\frac{1}{2 \sqrt{2}} {t a n}^{- 1} (- \infty) - 0

\frac{1}{2 \sqrt{2}} (- \frac{Π}{2})

a n s i s [\frac{1}{2 \sqrt{2}} {t a n}^{- 1} (\frac{x - \frac{1}{x}}{\sqrt{2}}) - \frac{1}{4 \sqrt{2}} l n ∣ \frac{x + \frac{1}{x} - \sqrt{2}}{x + \frac{1}{x} + \sqrt{2}} ∣ + \frac{Π}{4 \sqrt{2}}