0-t-1-t-2-dt-FAILED-TO-CALCULATE-

Question Number 168743 by mehdiAz last updated on 16/Apr/22

\int_{0}^{\infty} \frac{\sqrt{t}}{1 + t^{2}} d t

FAILED TO CALCULATE

Commented by GalaxyBills last updated on 17/Apr/22

\int_{0}^{\infty} \frac{\sqrt{t}}{1 + t^{2}} d t

FAILED TO CALCULATE

S o l u t i o n b y A w u d u l a i

l e t t = u^{2} \to d t = 2 u d u

\int_{0}^{\infty} \frac{2 u \sqrt{u^{2}}}{1 + u^{4}} d u \to \int_{0}^{\infty} \frac{2 u^{2}}{1 + u^{4}} d u

l e t z = u^{4} \overset{u = z^{\frac{1}{4}}}{\to} d z = 4 u^{3} \Rightarrow d u = \frac{d z}{4 u^{3}}

\int_{0}^{\infty} \frac{2 u^{2} . u^{- 3}}{4 (1 + z)} d z \to \frac{1}{2} \int_{0}^{\infty} \frac{z^{- \frac{1}{4}}}{(1 + z)} d z

u s i n g \int_{0}^{\infty} \frac{x^{m - 1}}{{(1 + x)}^{m + n}} d x \to Γ (m) Γ (n)

m = \frac{3}{4} a m d n = \frac{1}{4}

\frac{1}{2} Γ (\frac{3}{4}) Γ (\frac{1}{4}) \to \frac{1}{2} (\frac{π}{s i n (\frac{π}{4})}) = \frac{π \sqrt{2}}{2} = \frac{2 π}{2 \sqrt{2}} = \frac{π}{\sqrt{2}}

Kudos to my Mentors: Prempeh (Euclid)

and to Daniel (Small Einstein)

Answered by Mathspace last updated on 17/Apr/22

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

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

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

= 2 \int_{0}^{\infty} \frac{z^{\frac{1}{2}}}{1 + z} \frac{1}{4} z^{\frac{1}{4} - 1} d z

= \frac{1}{2} \int_{0}^{\infty} \frac{z^{\frac{3}{4} - 1}}{1 + z} d z = \frac{1}{2} \times \frac{π}{s i n (\frac{3 π}{4})}

= \frac{π}{2} . \frac{1}{\frac{1}{\sqrt{2}}} = \frac{π \sqrt{2}}{2} = \frac{π}{\sqrt{2}}