advanced-calculus-prove-that-0-pi-cos-tan-x-cot-x-dx-pi-e-2-

Question Number 129684 by mnjuly1970 last updated on 17/Jan/21

\dots a d v a n c e d c a l c u l u s \dots

p r o v e t h a t :

\int_{0}^{π} c o s (t a n (x) - c o t (x)) d x = \frac{π}{e^{2}}

Answered by mindispower last updated on 17/Jan/21

= 2 \int_{0}^{\frac{π}{2}} c o s (t g (x) - c o t (x)) d x

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

= R e \int_{- \infty}^{\infty} \frac{e^{i (t - \frac{1}{t})}}{1 + t^{2}}

t = {r e}^{i ϵ}, ϵ \in [0, π [

e^{i ({r e}^{i ϵ} - \frac{e^{- i ϵ}}{r})} = e^{i (t - \frac{1}{t})},

e^{i} ({r e}^{i ϵ} - \frac{e^{- i ϵ}}{r}) \to 0, r \to \infty

w e h n r \to 0

r \to 0^{+}

e^{i ({r e}^{i ϵ} + i \frac{s i n (ϵ) - c o s (ϵ)}{r})} \to 0

\Rightarrow \int_{- \infty}^{+ \infty} \frac{e^{i (t - \frac{1}{t})}}{1 + t^{2}} d t = 2 i π . (R e s (\frac{e^{i (t - \frac{1}{t})}}{1 + t^{2}}, t = i)

= 2 i π R e (\frac{e^{i (i - \frac{1}{i})}}{2 i}) = 2 i π . (\frac{e^{- 2}}{2 i}) = \frac{π}{e^{2}}

2 \int_{0}^{\infty} \frac{c o s (t - \frac{1}{t})}{1 + t^{2}} d t = R e (\frac{π}{e^{2}}) = \frac{π}{e^{2}}

Answered by mathmax by abdo last updated on 17/Jan/21

I = \int_{0}^{\frac{π}{2}} \cos (tanx - \frac{1}{tanx}) dx + \int_{\frac{π}{2}}^{π} \cos (tanx - \frac{1}{tanx}) dx (\to x = \frac{π}{2} + t)

= 2 \int_{0}^{\frac{π}{2}} \cos (tanx - \frac{1}{tanx}) dx =_{tanx = t} 2 \int_{0}^{\infty} \cos (t - \frac{1}{t}) \frac{dt}{t^{2} + 1}

= \int_{- \infty}^{+ \infty} \frac{\cos (t - \frac{1}{t})}{t^{2} + 1} dt = Re (\int_{- \infty}^{+ \infty} \frac{e^{i (t - \frac{1}{t})}}{t^{2} + 1} dt) let

φ (z) = \frac{e^{i (z - \frac{1}{z})}}{z^{2} + 1} \Rightarrow φ (z) = \frac{e^{i (z - \frac{1}{z})}}{(z - i) (z + i)} residus [theorem give

\int_{- \infty}^{+ \infty} φ (z) dz = 2 i π Res (φ, i) = 2 i π \times \frac{e^{i (i - \frac{1}{i})}}{2 i} = π e^{- 1 - 1} = \frac{π}{e^{2}} \Rightarrow

★ I = \frac{π}{e^{2}} ★

Answered by mnjuly1970 last updated on 20/Jan/21