calculate-0-dx-x-4-1-2-

Question Number 126493 by mathmax by abdo last updated on 21/Dec/20

calculate \int_{0}^{\infty} \frac{dx}{{(x^{4} + 1)}^{2}}

Answered by Dwaipayan Shikari last updated on 21/Dec/20

\int_{0}^{\infty} \frac{d x}{{(x^{4} + 1)}^{2}} x^{4} = t \Rightarrow 4 x^{3} = \frac{d t}{d x}

= \frac{1}{4} \int_{0}^{\infty} \frac{t^{- \frac{3}{4}}}{{(t + 1)}^{2}} d t \frac{t}{t + 1} = u

= \frac{1}{4} \int_{0}^{1} {(\frac{u}{1 - u})}^{- \frac{3}{4}} d u = \frac{1}{4} β (\frac{1}{4}, \frac{7}{4}) = \frac{3 Γ (\frac{1}{4}) Γ (\frac{3}{4})}{16} = \frac{3 π}{8 \sqrt{2}}

Answered by mathmax by abdo last updated on 21/Dec/20

let f (a) = \int_{0}^{\infty} \frac{dx}{x^{4} + a^{4}} with a > 0 \Rightarrow f^{^{'}} (a) = - \int_{0}^{\infty} \frac{{4 a}^{3}}{{(x^{4} + a^{4})}^{2}} dx \Rightarrow

f^{^{'}} (1) = - 4 \int_{0}^{\infty} \frac{dx}{{(x^{4} + 1)}^{2}} \Rightarrow \int_{0}^{\infty} \frac{dx}{{(x^{4} + 1)}^{2}} = - \frac{1}{4} f^{^{'}} (1) let explicit f (a)

f (a) =_{x = at} \int_{0}^{\infty} \frac{adt}{a^{4} (t^{4} + 1)} = \frac{1}{a^{3}} \int_{0}^{\infty} \frac{dt}{1 + t^{4}} =_{t = z^{\frac{1}{4}}}

= \frac{1}{{4 a}^{3}} \int_{0}^{\infty} \frac{1}{1 + z} z^{\frac{1}{4} - 1} dz = \frac{1}{{4 a}^{3}} \times \frac{π}{\sin (\frac{π}{4})} = \frac{π}{{4 a}^{3} \times \frac{\sqrt{2}}{2}} = \frac{π}{2 \sqrt{2} a^{3}} \Rightarrow

f (a) = \frac{π}{2 \sqrt{2}} a^{- 3} \Rightarrow f^{^{'}} (a) = \frac{π}{2 \sqrt{2}} (- 3) a^{- 4} \Rightarrow f^{^{'}} (1) = \frac{- 3 π}{2 \sqrt{2}} \Rightarrow

\int_{0}^{\infty} \frac{dx}{{(x^{4} + 1)}^{2}} = - \frac{1}{4} \times \frac{- 3 π}{2 \sqrt{2}} = \frac{3 π}{8 \sqrt{2}} = \frac{3 π \sqrt{2}}{16}