2 1 Thus the Cauchy’s principal value of dx ln 2  1 x Example 7 dx Show that 2 x 4 3 i). does not converge ii). does not have a principal value Solution dx The integral is improper x = 0 is a singularity on the interval [-2 x ≤ 2 x 4 3 3] ≤ dx dx dx   Now 2 - 2 x 0 x x 4 4 4 0 3 3 91 0 3 1 1 3 3 3 3 x x - 2 0 1 1  24 81 does not converge The Cauchy’s principal value of   3 dx dx - 2 x x dx  where 0 4 4  3 2 x 4 1 1 3 3 - 2  3 3 x x -  3 0 where = 3 1 3 1 1  1 3  3  81 24
Thus the Cauchy’s principal value of 2 1 1 x dx  ln 2 Example 7 Show that 3 2 dx x 4 i). does not converge ii). does not have a principal value Solution The integral 3 2 dx x 4 is improper x = 0 is a singularity on the interval [-2 ≤ x ≤ 3] Now 3 2 dx x 4  0 - 2 dx x 4  3 0 dx x 4 91 1 3 x 3 0 - 2 1 3 x 3 3 0  1 24 1 81 does not converge The Cauchy’s principal value of 3 2 dx x 4   - 2 dx x 4  3 dx x 4 where  0 1 3 x 3 -  - 2 1 3 x 3 3   1 3  0 = 3 1 24 1 81 1 3  3 where
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