MATH502_HW5 - Serge Ballif(1 Evaluate the integral MATH 502...

Info icon This preview shows pages 1–2. Sign up to view the full content.

Serge Ballif MATH 502 Homework 5 February 22, 2008 (1) Evaluate the integral Z -∞ x 2 cos x 4 + x 4 d x by means of contour integration. Be sure to justify carefully all the steps in your argument. To compute this integral, we will consider the function h ( z ) = z 2 e iz 4+ z 4 , which is bounded in the upper half plane and has real part equal to the integrand in our problem. We will integrate over the curve γ 1 and γ 2 as pictured below. R - R 1 + i - 1 + i γ 2 γ 1 Call the entire path Γ. Then by the residue theorem 1 2 πi Z Γ f ( z ) d z = Res( h, 1 + i ) + Res( h, - 1 + i ) . Since | e iz | ≤ 1 in the upper half plane and tends to 0 almost everywhere as R , we see that R γ 2 z 2 e z 4+ z 4 d z tends to 0 as R → ∞ by the dominated convergence theorem. Therefore, by the residue theorem Z Γ z 2 e z 4 + z 4 d z = Z γ 1 z 2 e z 4 + z 4 d z = 2 πi (Res( h, 1 + i ) + Res( h, - 1 + i )) . We are finished when we compute this, because R -∞ x 2 cos x 4+ x 4 d x = lim R →∞ R γ 1 z 2 e z 4+ z 4 d z To calculate the residues we note that z 4 + 4 = ( z - 1 - i )( z - 1 + i )( z + 1 - i )( z +1+ i ), so h ( z ) has simple poles at z = ± 1 ± i . We can compute the residues using the formula for simple poles: Res( h, 1 + i ) = (1+ i ) 2 e i (1+ i ) 4(1+ i ) 3 = (1 - i ) e - 1+ i / 8. Similarly Res( h, - 1 + i ) = ( - 1 - i ) e - 1 - i / 8. Thus Z -∞ x 3 cos x 4 + x 4 d x = Re 2 πi e - 1 ( ( e i - e - i ) + i ( - e i - e - i ) ) 8 !
Image of page 1

Info icon This preview has intentionally blurred sections. Sign up to view the full version.

Image of page 2
This is the end of the preview. Sign up to access the rest of the document.
  • Summer '08
  • JOHNRIDENER
  • Math, Methods of contour integration, dz, Serge Ballif

{[ snackBarMessage ]}

What students are saying

  • Left Quote Icon

    As a current student on this bumpy collegiate pathway, I stumbled upon Course Hero, where I can find study resources for nearly all my courses, get online help from tutors 24/7, and even share my old projects, papers, and lecture notes with other students.

    Student Picture

    Kiran Temple University Fox School of Business ‘17, Course Hero Intern

  • Left Quote Icon

    I cannot even describe how much Course Hero helped me this summer. It’s truly become something I can always rely on and help me. In the end, I was not only able to survive summer classes, but I was able to thrive thanks to Course Hero.

    Student Picture

    Dana University of Pennsylvania ‘17, Course Hero Intern

  • Left Quote Icon

    The ability to access any university’s resources through Course Hero proved invaluable in my case. I was behind on Tulane coursework and actually used UCLA’s materials to help me move forward and get everything together on time.

    Student Picture

    Jill Tulane University ‘16, Course Hero Intern