Prove that (with *a*>0).

First, we break up the improper integral at the *x*=0 singularity:

.

Next, consider the u substitution ; then we see , and so we have two different solutions for *x* as a function of *u*. Using the quadratic formula, we find ; since for all real *u* when *a* is positive, we see and .

Thus, for our first integral, *x*<0, we have

,

so

;

and we see as , and as , and so, making the substitution,

.

Similarly, for our second integral, *x*>0, and we have

,

so

;

and we see as , and as , and so, making the substitution,

.

Combining these,

.

Q.E.D.

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June 7, 2010 at 12:12 am |

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