Mathematics Tripos Papers

  • Part IA
  • Part IB
  • Part II
  • FAQ

3.I.8E

Further Complex Methods | Part II, 2006

Show that, for b≠0b \neq 0b=0,

P∫0∞cos⁡uu2−b2du=−π2bsin⁡b\mathcal{P} \int_{0}^{\infty} \frac{\cos u}{u^{2}-b^{2}} d u=-\frac{\pi}{2 b} \sin bP∫0∞​u2−b2cosu​du=−2bπ​sinb

where P\mathcal{P}P denotes the Cauchy principal value.

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