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integration by residues

\displaystyle I = \int_{0}^{\infty} \frac{t^\frac{3}{4}}{t^2-1} \ dx

the theorem says this is equal to

-(\pi e^{-\frac{3\pi i}{4}} cosec(\frac{3\pi}{4}))S-(\pi cot(\frac{3\pi}{4}))T[br]

with S is the sum of the residues of the function g(z):

\frac{1}{t^{2}-1} exp (\frac {3}{4} Log_{2\pi}(z)) [br]

C_{2\pi}

and T is the sum of the residues of h(z)

\frac{1}{t^{2}-1} exp (\frac {3}{4} Log (z)) [br]

on positive real axis

now only zero in

C_{2\pi}

is -1

and I get

res (g(z),-1) to be

\frac{-1}{2}e^{\frac{3\pi i}{4}}

res (h(z),1) to be 1/2

so putting that in the result from the theorem I get

-(\pi (e^{\frac{-3\pi i}{4}})(\sqrt{2}) (\frac{-1}{2}e^{\frac{3\pi i}{4}}) +(\pi)(1/2) [br][br]

which gives

\pi ( \frac{1}{2}+\frac{\sqrt(2)}{2})[br]

which all looks good but when I checked this with wolfram is says integral diverges?!!

I normally believe wolfram but as this is a question from a past paper I got to think it has a value.

any suggestions as to my answer or wolfram's?

(edited 7 years ago)

Reply 1

mathz

not sure what is wrong with wolfram.

this site evaluates integral, and its the answer I found so even better :smile:

http://www.mathportal.org/calculators/calculus/integral-calculator.php

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integration by residues

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