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Question Number 86447 by Chi Mes Try last updated on 28/Mar/20

Commented by mathmax by abdo last updated on 29/Mar/20
$let f(t) =∫_0 ^∞ ((arctan(t(√(x^2 +a^2 ))))/((1+x^2 )(√(x^2 +a^2 ))))dx with t>0 f^′ (t)=∫_0 ^∞ (1/((1+x^2 )(1+t^2 (x^2 +a^2 )))dx =∫_0 ^∞ (dx/((1+x^2 )(1+t^2 x^2 +a^2 t^2 ))) =∫_0 ^∞ (dx/((x^2 +1)(t^2 x^2 +a^2 t^2 +1))) =(1/t^2 )∫_0 ^∞ (dx/((x^2 +1)(x^2 +((a^2 t^2 +1)/t^2 )))) ⇒ 2t^2 f^′ (t) =∫_(−∞) ^(+∞) (dx/((x−i)(x+i)(x−i((√(a^2 t^2 +1))/t))(x+i((√(a^2 t^2 +1))/t)))) let ϕ(z) =(1/((z^2 +1)(z^2 +((a^2 t^2 +1)/t^2 )))) ∫_(−∞) ^(+∞) ϕ(z)dz =2iπ {Res(ϕ,i)+Res(ϕ,i((√(a^2 t^2 +1))/t))} Res(ϕ,i) =(1/(2i(−1+((a^2 t^2 +1)/t^2 )))) =(t^2 /(2i(a^2 t^2 +1−t^2 ))) =(t^2 /(2i{(a^2 −1)t^2 +1)})) Res(ϕ,i((√(a^2 t^2 +1))/t)) =(1/(2i((√(a^2 t^2 +1))/t)( −((a^2 t^2 +1)/t^2 )+1))) =(t^3 /(2i(√(a^2 t^2 +1))(t^2 −a^2 t^2 −1))) =(t^3 /(2i(√(a^2 t^2 +1))((1−a^2 )t^2 −1))) ⇒ ∫_(−∞) ^(+∞) ϕ(z)dz =2iπ{ (t^2 /(2i{(a^2 −1)t^2 +1})) +(t^3 /(2i(√(a^2 t^2 +1)){(1−a^2 )t^2 −1}))} =((πt^2 )/((a^2 −1)t^2 +1)) +((πt^3 )/((√(a^2 t^2 +1)){(1−a^2 )t^2 −1}))=2t^2 f^′ (t) ⇒ f^′ (t) =(π/(2{(a^2 −1)t^2 +1})) +((πt)/(2(√(a^2 t^2 +1)){(1−a^2 )u^2 −1})) ⇒ f(t) =∫_0 ^t ((πdu)/(2{(a^2 −1)u^2 +1})) +(π/2)∫_0 ^t ((udu)/((√(a^2 u^2 +1)){(1−a^2 )u^2 −1})) +c c=f(0)=0 I =f(1) =(π/2)∫_0 ^1 (du/((a^2 −1)u^2 +1)) +(π/2)∫_0 ^1 ((udu)/((√(a^2 u^2 +1)){(1−a^2 )u^2 −1})) ....be continued....$
$l e t f (t) = \int_{0}^{\infty} \frac{a r c t a n (t \sqrt{x^{2} + a^{2}})}{(1 + x^{2}) \sqrt{x^{2} + a^{2}}} d x w i t h t > 0$ $f^{^{'}} (t) = \int_{0}^{\infty} \frac{1}{(1 + x^{2}) (1 + t^{2} (x^{2} + a^{2})} d x$ $= \int_{0}^{\infty} \frac{d x}{(1 + x^{2}) (1 + t^{2} x^{2} + a^{2} t^{2})} = \int_{0}^{\infty} \frac{d x}{(x^{2} + 1) (t^{2} x^{2} + a^{2} t^{2} + 1)}$ $= \frac{1}{t^{2}} \int_{0}^{\infty} \frac{d x}{(x^{2} + 1) (x^{2} + \frac{a^{2} t^{2} + 1}{t^{2}})} \Rightarrow$ $2 t^{2} f^{^{'}} (t) = \int_{- \infty}^{+ \infty} \frac{d x}{(x - i) (x + i) (x - i \frac{\sqrt{a^{2} t^{2} + 1}}{t}) (x + i \frac{\sqrt{a^{2} t^{2} + 1}}{t})}$ $l e t φ (z) = \frac{1}{(z^{2} + 1) (z^{2} + \frac{a^{2} t^{2} + 1}{t^{2}})}$ $\int_{- \infty}^{+ \infty} φ (z) d z = 2 i π {R e s (φ, i) + R e s (φ, i \frac{\sqrt{a^{2} t^{2} + 1}}{t})}$ $R e s (φ, i) = \frac{1}{2 i (- 1 + \frac{a^{2} t^{2} + 1}{t^{2}})} = \frac{t^{2}}{2 i (a^{2} t^{2} + 1 - t^{2})} = \frac{t^{2}}{2 i {(a^{2} - 1) t^{2} + 1)}}$ $R e s (φ, i \frac{\sqrt{a^{2} t^{2} + 1}}{t}) = \frac{1}{2 i \frac{\sqrt{a^{2} t^{2} + 1}}{t} (- \frac{a^{2} t^{2} + 1}{t^{2}} + 1)}$ $= \frac{t^{3}}{2 i \sqrt{a^{2} t^{2} + 1} (t^{2} - a^{2} t^{2} - 1)} = \frac{t^{3}}{2 i \sqrt{a^{2} t^{2} + 1} ((1 - a^{2}) t^{2} - 1)} \Rightarrow$ $\int_{- \infty}^{+ \infty} φ (z) d z = 2 i π {\frac{t^{2}}{2 i {(a^{2} - 1) t^{2} + 1}} + \frac{t^{3}}{2 i \sqrt{a^{2} t^{2} + 1} {(1 - a^{2}) t^{2} - 1}}}$ $= \frac{π t^{2}}{(a^{2} - 1) t^{2} + 1} + \frac{π t^{3}}{\sqrt{a^{2} t^{2} + 1} {(1 - a^{2}) t^{2} - 1}} = 2 t^{2} f^{^{'}} (t) \Rightarrow$ $f^{^{'}} (t) = \frac{π}{2 {(a^{2} - 1) t^{2} + 1}} + \frac{π t}{2 \sqrt{a^{2} t^{2} + 1} {(1 - a^{2}) u^{2} - 1}} \Rightarrow$ $f (t) = \int_{0}^{t} \frac{π d u}{2 {(a^{2} - 1) u^{2} + 1}} + \frac{π}{2} \int_{0}^{t} \frac{u d u}{\sqrt{a^{2} u^{2} + 1} {(1 - a^{2}) u^{2} - 1}} + c$ $c = f (0) = 0$ $I = f (1) = \frac{π}{2} \int_{0}^{1} \frac{d u}{(a^{2} - 1) u^{2} + 1} + \frac{π}{2} \int_{0}^{1} \frac{u d u}{\sqrt{a^{2} u^{2} + 1} {(1 - a^{2}) u^{2} - 1}}$ $. . . . b e c o n t i n u e d . . . .$
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