calculatef-a-0-ln-1-at-2-1-t-4-dt-with-a-gt-0-2-find-the-value-of-0-ln-3-t-2-1-t-4-dt-

Question Number 51834 by Abdo msup. last updated on 31/Dec/18

c a l c u l a t e f (a) = \int_{0}^{\infty} \frac{l n (1 + {a t}^{2})}{1 + t^{4}} d t w i t h a > 0 .

2) f i n d t h e v a l u e o f \int_{0}^{\infty} \frac{l n (3 + t^{2})}{1 + t^{4}} d t .

Commented by Abdo msup. last updated on 31/Dec/18

1) w e h a v e f^{^{'}} (a) = \int_{0}^{\infty} \frac{t^{2}}{(1 + {a t}^{2}) (t^{4} + 1)} d t

= \frac{1}{2} \int_{- \infty}^{+ \infty} \frac{t^{2}}{({a t}^{2} + 1) (t^{4} + 1)} d t l e t

φ (z) = \frac{z^{2}}{({a z}^{2} + 1) (z^{4} + 1)} w e h a v e

φ (z) = \frac{z^{2}}{(\sqrt{a} z - i) (\sqrt{a} z + i) (z^{2} - i) (z^{2} + i)}

= \frac{z^{2}}{a (z - \frac{i}{\sqrt{a}}) (z + \frac{i}{\sqrt{a}}) (z - e^{\frac{i π}{4}}) (z + e^{\frac{i π}{4}}) (z - e^{- \frac{i π}{4}}) (z + e^{- \frac{i π}{4}})}

t h e p o l e s o f φ a r e \bar{+} \frac{i}{\sqrt{a}} a n d \bar{+} e^{\frac{i π}{4}} a n d \bar{+} e^{- \frac{i π}{4}} (a l l s i m p l e s)

r e s i d u s t b e o r e m g i v e

\int_{- \infty}^{+ \infty} φ (z) d z = 2 i π {R e s (φ, \frac{i}{\sqrt{a}}) + R e s (φ, e^{\frac{i π}{4}}) + R e s (φ, - e^{- \frac{i π}{4}})}

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

R e s (φ, e^{\frac{i π}{4}}) = \frac{i}{(a i + 1) (2 e^{\frac{i π}{4}}) (2 i)} = \frac{e^{- \frac{i π}{4}}}{4 (1 + a i)}

R e s (φ, - e^{- \frac{i π}{4}}) = \frac{- i}{(1 - a i) (- 2 e^{- \frac{i π}{4}}) (- 2 i)}

= - \frac{e^{\frac{i π}{4}}}{4 (1 - a i)} \Rightarrow

\int_{- \infty}^{+ \infty} φ (z) d z = 2 i π {\frac{i \sqrt{a}}{2 (1 + a^{2})} - \frac{1}{4} (2 i I m (\frac{e^{\frac{i π}{4}}}{1 - a i})}

= \frac{- π \sqrt{a}}{1 + a^{2}} - π I m (\frac{e^{\frac{i π}{4}}}{1 - a i}) b u t

\frac{e^{\frac{i π}{4}}}{1 - a i} = \frac{e^{\frac{i π}{4}} (1 + a i)}{1 + a^{2}} = \frac{(\frac{1}{\sqrt{2}} + \frac{i}{\sqrt{2}}) (1 + a i)}{1 + a^{2}}

= \frac{1}{\sqrt{2}} \frac{(1 + i) (1 + a i)}{1 + a^{2}} = \frac{1 + a i + i - a}{\sqrt{2} (1 + a^{2})} = \frac{1 - a + i (1 + a)}{\sqrt{2} (1 + a^{2})} \Rightarrow

\int_{- \infty}^{+ \infty} φ (z) d z = \frac{- π \sqrt{a}}{1 + a^{2}} - π \frac{1 + a}{\sqrt{2} (1 + a^{2})} \Rightarrow

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

f (a) = - \frac{π}{2} \int \frac{\sqrt{a} d a}{1 + a^{2}} - \frac{π}{2 \sqrt{2}} \int \frac{1 + a}{1 + a^{2}} d a + c