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Mathematics 16 Online
OpenStudy (anonymous):

Prove that the integral from 0 to infinity sinx/x =pie/2

OpenStudy (anonymous):

\[\int_0^{\infty} \frac{\sin {x}}{x}\]

OpenStudy (anonymous):

hmm did you try by parts ??

OpenStudy (anonymous):

integration by parts

OpenStudy (anonymous):

no. I havent

OpenStudy (anonymous):

i am confused about the x in denominator i dunno how to get rid of it

OpenStudy (anonymous):

i think it's Si(x) function not sure lemme check

OpenStudy (anonymous):

http://www.wolframalpha.com/input/?i=sinx%2Fx+dx+ yeah it is Si(x)

OpenStudy (anonymous):

Hmm i dunno about them sorry I won;t be able to help you

OpenStudy (anonymous):

this site http://tutorial.math.lamar.edu/Classes/CalcII/IntegrationStrategy.aspx says that the integral can never be solved except with an algebra system

OpenStudy (anonymous):

algebra system hmm sounds advanced yet interesting

OpenStudy (anonymous):

certainly

OpenStudy (anonymous):

can working be shown as to how you can arrive at the answer?

OpenStudy (lalaly):

did u take fourier series?

OpenStudy (anonymous):

I seriously have no idea on how to do this, i have to study much more to get on something like this

OpenStudy (anonymous):

lalaly I didnt but could u give me insight on that?

OpenStudy (zarkon):

you can write it as a double integral and then switch the order of integration... \[\int\limits_{0}^{\infty}\frac{\sin(x)}{x}dx=\int\limits_{0}^{\infty}\int\limits_{0}^{\infty}e^{-xt}\sin(x)dtdx\]

OpenStudy (anonymous):

thanx Ishaan I will await your reply

OpenStudy (anonymous):

Zarkon could you please finish it up for me?

OpenStudy (zarkon):

I really don't want to ;)...switch the order of integration...then use parts on the first integral...you will end up with a standard inverse trig integral

OpenStudy (anonymous):

thanx Zarkon

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