Sunday, September 15, 2019

calculus - Fresnel integral intlimitsi0nftysin(x2)dx calculation



I'm trying to calculate the improper Fresnel integral 0sin(x2)dx calculation.
It uses several substitutions. There's one substitution that is not clear for me.



I could not understand how to get the right side from the left one. What subtitution is done here?



0v21+v4dv=1201+u21+u4du.







Fresnel integral calculation:



In the beginning put x2=t and then: 0sin(x2)dx=120sinttdt



Then changing variable in Euler-Poisson integral we have: 2π0etu2du=1t



The next step is to put this integral instead of 1t.

0sin(x2)dx=1π0sin(t)0 etu2dudt=1π00sin(t)etu2dtdu
And the inner integral 0sin(t)etu2dt is equal to 11+u4.



The next calculation: 0du1+u4=0v2dv1+v4=1201+u21+u4du=120d(u1u)u2+1u2
=12ds2+s2=12arctans2|=π22



In this calculation the Dirichle's test is needed to check the integral 0sinttdt convergence. It's needed also to substantiate the reversing the order of integration (dudt=dtdu). All these integrals exist in a Lebesgue sense, and Tonelli theorem justifies reversing the order of integration.



The final result is 1ππ22=12π2


Answer




Well, if one puts v=1u then:
I=0v21+v4dv=011+u4du
So summing up the two integrals from above gives:
2I=01+u21+u4du


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