## Monday Math 121

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.