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P. 30

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                                                                                          Unit 1: Bessel s Functions




          Solution: From example above, we have                                                 Notes

                           1
                                x
                     J (x) =   cos( sin )dx
                     0
                            0
                                1
                ax
                             ax
               e  j (bx )dx  e    cos(bx  sin )d  dx
                  0
                                x
              0            0     0
                           1     ax
                               e  cos(bx  sin )dx d
                            0  0

                           1     ax e  ( i bx sin )  e  ( i bx sin )
                               e                 dx d
                                         2
                            0  0


                           1      (a ib sin )x  (    i sin )x
                                               b
                                            a
                                 e        e         dx d
                          2
                             0  0
                                               ib
                                             a
                           1    e  (a ib  sin )x  e  (     sin )x
                                                      d
                          2     (a ib sin )  (a ib sin )
                             0                      0
                           1      1         1
                                                 d
                          2    a ib sin  a ib sin
                             0
                           1     2  d
                                  a

                          2   a 2   b  2 sin  2
                             0
                              /2      2
                            a     cosec  d
                          2.     2   2    2
                                b   a  cosec
                              0
                              /2       2
                            a      cosec  d
                          2.
                                (a  2  b 2  ) a 2  cot 2
                              0
                                                       /2
                            a      1       1   a cot
                          2.            cot
                              a  {(a 2  b 2 )}  {(a 2  b 2  )}
                                                      0

                               2         1     1
                                      cot 0 cot
                             {(a 2  b 2 )}

                              1

                            {(a 2  b 2 )}




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