1-find-1-dx-x-2-i-and-1-dx-x-2-i-i-1-2-find-the-value-of-1-dx-x-4-1-

Question Number 96373 by mathmax by abdo last updated on 01/Jun/20

1 . find \int_{1}^{+ \infty} \frac{dx}{x^{2} - i} and \int_{1}^{+ \infty} \frac{dx}{x^{2} + i} (i = \sqrt{- 1})

2 . find the value of \int_{1}^{+ \infty} \frac{dx}{x^{4} + 1}

Answered by Sourav mridha last updated on 01/Jun/20

l e t I = \int_{1}^{+ \infty} \frac{d x}{x^{2} - i} a n d J = \int_{1}^{+ \infty} \frac{d x}{x^{2} + i}

n o w, I + J = \int_{1}^{+ \infty} \frac{2 x^{2}}{x^{4} + 1} d x \dots . . (i)

= \int_{1}^{+ \infty} \frac{x^{2} + 1}{x^{4} + 1} d x + \int_{1}^{\infty} \frac{x^{2} - 1}{x^{4} + 1} d x

\dots \dots . . (I_{1}) \dots \dots \dots . (I_{2})

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

= \frac{1}{\sqrt{2}} {[\tan^{- 1} (\frac{x - \frac{1}{x}}{\sqrt{2}})]}_{1}^{+ \infty} + \frac{1}{2 \sqrt{2}} {[l n (\frac{x + \frac{1}{x} - \sqrt{2}}{x + \frac{1}{x} + \sqrt{2}})]}_{1}^{+ \infty}

= \frac{π}{2 \sqrt{2}} - \frac{1}{2 \sqrt{2}} l n (3 - \sqrt{2})

n o w I - J = \frac{1}{2 i} \int_{1}^{+ \infty} \frac{d x}{x^{4} + 1} \dots \dots (i i)

= \frac{1}{4 i} [\int_{1}^{+ \infty} \frac{x^{2} + 1}{x^{4} + 1} d x - \int_{1}^{+ \infty} \frac{x^{2} - 1}{x^{4} + 1} d x]

= \frac{1}{4 i} [I_{1} - I_{2}] = - \frac{i}{4} [\frac{π}{2 \sqrt{2}} + \frac{1}{2 \sqrt{2}} l n (3 - \sqrt{2})]

n o w (i) + (i i)

I = [\frac{π}{4 \sqrt{2}} - \frac{1}{4 \sqrt{2}} l n (3 - \sqrt{2})] - i [\frac{π}{16 \sqrt{2}} + \frac{1}{16 \sqrt{2}} l n (3 - \sqrt{2})]

a n d (i) - (i i)

J = I^{*} = c o m p l e x c o n j u g a t e o f I .

n o w f r o m (i i) \dots

\int_{1}^{+ \infty} \frac{d x}{x^{4} + 1} = 2 i (I - J)

= [\frac{π}{4 \sqrt{2}} + \frac{1}{4 \sqrt{2}} l n (3 - \sqrt{2})] .

v e r y n i c e p r o b l e m \dots {i t}^{'} s r e a l l y

i n t e r e s t i n g .

Commented by abdomathmax last updated on 01/Jun/20

thanks sir

Answered by abdomathmax last updated on 01/Jun/20

1 . \int_{1}^{+ \infty} \frac{dx}{x^{2} - i} = \int_{1}^{+ \infty} \frac{dx}{(x - \sqrt{i}) (x + \sqrt{i})}

= \frac{1}{2 \sqrt{i}} \int_{1}^{+ \infty} (\frac{1}{x - \sqrt{i}} - \frac{1}{x + \sqrt{i}}) dx

= \frac{1}{2 \sqrt{i}} {[\ln (\frac{x - \sqrt{i}}{x + \sqrt{i}})]}_{1}^{+ \infty} = \frac{1}{2 \sqrt{i}} (- \ln (\frac{1 - \sqrt{i}}{1 + \sqrt{i}}))

= \frac{1}{2 \sqrt{i}} (\ln (1 + \sqrt{i}) - \ln (1 - \sqrt{i})) but

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

e^{iarctan (\frac{1}{\sqrt{2} (1 + \frac{1}{\sqrt{2}}})} = \sqrt{\frac{3}{2} + \sqrt{2} + \frac{1}{2}} e^{iarctan (\frac{1}{1 + \sqrt{2}})}

= \sqrt{2 + \sqrt{2}} e^{iarctan (\frac{1}{1 + \sqrt{2}})} \Rightarrow

\ln (1 + \sqrt{i}) = \frac{1}{2} \ln (2 + \sqrt{2}) + iarctan (\frac{1}{1 + \sqrt{2}})

\ln (1 - \sqrt{i}) = conj (. .) = \frac{1}{2} \ln (2 + \sqrt{2}) - iarctan (\frac{1}{1 + \sqrt{2}})

\Rightarrow \frac{1}{2 \sqrt{i}} (\ln (1 + \sqrt{i}) - \ln (1 - \sqrt{i}))

= \frac{1}{2 \sqrt{i}} {2 i \arctan (\frac{1}{1 + \sqrt{2}})} = \sqrt{i} \arctan (\sqrt{2} - 1)

= \sqrt{i} (\frac{π}{8}) = \frac{π}{8} e^{\frac{i π}{4}}

\int_{1}^{+ \infty} \frac{dx}{x^{2} + i} = conj (\int_{1}^{+ \infty} \frac{dx}{x^{2} - i}) = \frac{π}{8} e^{- \frac{i π}{4}}

2 . \int_{1}^{+ \infty} \frac{dx}{x^{4} + 1} = \int_{1}^{+ \infty} \frac{dx}{(x^{2} - i) (x^{2} + i)}

= \frac{1}{2 i} \int_{1}^{+ \infty} (\frac{1}{x^{2} - i} - \frac{1}{x^{2} + i}) dx

= \frac{1}{2 i} (\frac{π}{8} e^{\frac{i π}{4}} - \frac{π}{8} e^{- \frac{i π}{4}})

= \frac{π}{8} \sin (\frac{π}{4}) = \frac{π}{8} \times \frac{\sqrt{2}}{2} = \frac{π \sqrt{2}}{16}