calculate-0-2pi-dt-p-cost-with-p-gt-1-

Question Number 37348 by math khazana by abdo last updated on 12/Jun/18

c a l c u l a t e \int_{0}^{2 π} \frac{d t}{p + c o s t} w i t h p > 1

Commented by prof Abdo imad last updated on 13/Jun/18

I = \int_{0}^{π} \frac{d t}{p + c o s t} + \int_{π}^{2 π} \frac{d t}{p + c o s t} = K + H

c h a n g e m e n t t a n (\frac{t}{2}) = x g i v e

K = \int_{0}^{\infty} \frac{1}{p + \frac{1 - x^{2}}{1 + x^{2}}} \frac{2 d x}{1 + x^{2}} = 2 \int_{0}^{\infty} \frac{d x}{p + {p x}^{2} + 1 - x^{2}}

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

c h a n g . \sqrt{\frac{p - 1}{p + 1}} x = u g i v e

K = \frac{2}{p + 1} \int_{0}^{+ \infty} \frac{1}{1 + u^{2}} \sqrt{\frac{p + 1}{p - 1}} d u

= \frac{2}{\sqrt{p^{2} - 1}} . \frac{π}{2} = \frac{π}{\sqrt{p^{2} - 1}} .

c h a n g e m e n t t = π + x g i v e

H = \int_{0}^{π} \frac{d x}{p - c o s x} =_{t a n (\frac{x}{2}) = t} \int_{0}^{\infty} \frac{1}{p - \frac{1 - t^{2}}{1 + t^{2}}} \frac{2 d t}{1 + t^{2}}

= \int_{0}^{\infty} \frac{2 d t}{p + {p t}^{2} - 1 + t^{2}} = \underset{x \to 0}{}

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

=_{\sqrt{\frac{p + 1}{p - 1}} t = u} \frac{2}{p - 1} \int_{0}^{\infty} \frac{1}{1 + u^{2}} \sqrt{\frac{p - 1}{p + 1}} d u

= \frac{2}{\sqrt{p^{2} - 1}} . \frac{π}{2} = \frac{π}{\sqrt{p^{2} - 1}} \Rightarrow

I = \frac{π}{\sqrt{p^{2} - 1}} + \frac{π}{\sqrt{p^{2} - 1}}

I = \frac{2 π}{\sqrt{p^{2} - 1}} w i t h p > 1 .

Commented by prof Abdo imad last updated on 13/Jun/18

R e s i d u s m e t h o d

c h a n g e m e n t e^{i t} = z g i v e

I = \int_{∣ z ∣= 1} \frac{1}{p + \frac{z + z^{- 1}}{2}} \frac{d z}{i z}

= \int_{∣ z ∣= 1} \frac{- 2 i d z}{z (2 p + z + z^{- 1})} = \int_{∣ z ∣= 1} \frac{- 2 i d z}{2 p z + z^{2} + 1}

= \int_{∣ z ∣= 1} \frac{- 2 i d z}{z^{2} + 2 p z + 1} l e t φ (z) = \frac{- 2 i}{z^{2} + 2 p z + 1}

p o l e s o f φ ?

Δ^{^{'}} = p^{2} - 1 \Rightarrow z_{1} = - p + \sqrt{p^{2} - 1}

z_{2} = - p - \sqrt{p^{2} - 1}

∣ z_{1} ∣ - 1 = p - \sqrt{p^{2} - 1} - 1 = p - 1 - \sqrt{p^{2} - 1}

{(p - 1)}^{2} - (p^{2} - 1) = p^{2} - 2 p + 1 - p^{2} + 1

= - 2 p + 2 = - 2 (p - 1) < 0 \Rightarrow∣ z_{1} ∣< 1

∣ z_{2} ∣ - 1 = p + \sqrt{p^{2} - 1} - 1 > 0 (t o e l i m i n a t e f r o m

R e s i d u s)

\int_{∣ z ∣= 1} φ (z) d z = 2 i π R e s (φ, z_{1})

R e s (φ, z_{1}) = \frac{- 2 i}{z_{1} - z_{2}} = \frac{- 2 i}{2 \sqrt{p^{2} - 1}} = \frac{- i}{\sqrt{p^{2} - 1}}

\int_{∣ z ∣= 1} φ (z) d z = 2 i π (\frac{- i}{\sqrt{p^{2} - 1}})

= \frac{2 π}{\sqrt{p^{2} - 1}} \Rightarrow

★ I = \frac{2 π}{\sqrt{p^{2} - 1}} ★