let-I-t-1-t-2-t-1-2-dt-find-value-of-I-

Question Number 53113 by maxmathsup by imad last updated on 17/Jan/19

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

f i n d v a l u e o f I .

Commented by maxmathsup by imad last updated on 18/Jan/19

w e h a v e t^{2} - t + 1 = {(t - \frac{1}{2})}^{2} + \frac{3}{4} s o c h a n g e m e n t t - \frac{1}{2} = \frac{\sqrt{3}}{2} x g i v e

I = \int_{- \infty}^{+ \infty} \frac{\frac{1}{2} + \frac{\sqrt{3}}{2} x + 1}{{(\frac{3}{4})}^{2} {(x^{2} + 1)}^{2}} \frac{\sqrt{3}}{2} d x = \frac{16}{9} \frac{\sqrt{3}}{4} \int_{- \infty}^{+ \infty} \frac{3 + \sqrt{3} x}{{(x^{2} + 1)}^{2}} d x

= \frac{4 \sqrt{3}}{9} \int_{- \infty}^{+ \infty} \frac{3 + \sqrt{3} x}{{(x^{2} + 1)}^{2}} d x l e t φ (z) = \frac{3 + \sqrt{3} z}{{(z^{2} + 1)}^{2}} w e h a v e

φ (z) = \frac{3 + \sqrt{3} z}{{(z - i)}^{2} {(z + i)}^{2}} s o t h e p l e s o f φ a r e \bar{+} i (d o u b l e s) a n d r e s i d u s t h e o r e m

g i v e \int_{- \infty}^{+ \infty} φ (z) d z = 2 i π R e s (φ, i)

R e s (φ, i) = {l i m}_{z \to i} \frac{1}{(2 - 1)!} {{(z - i)}^{2} φ (z)}^{(1)} = {l i m}_{z \to i} {\frac{3 + \sqrt{3} z}{{(z + i)}^{2}}}^{(1)}

= {l i m}_{z \to i} \frac{\sqrt{3} {(z + i)}^{2} - 2 (z + i) (3 + \sqrt{3} z)}{{(z + i)}^{4}}

= {l i m}_{z \to i} \frac{\sqrt{3} (z + i) - 2 (3 + \sqrt{3} z)}{{(z + i)}^{3}} = \frac{2 \sqrt{3} i - 2 (3 + \sqrt{3} i)}{- 8 i} = \frac{- 6}{- 8 i} = \frac{3}{4 i} \Rightarrow

\int_{- \infty}^{+ \infty} φ (z) d z = 2 i π \frac{3}{4 i} = \frac{3 π}{2} \Rightarrow I = \frac{4 \sqrt{3}}{9} \frac{3 π}{2} = \frac{2 π \sqrt{3}}{3} .