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.

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One Response to “Monday Math 121”

  1. Monday Math 122 « Twisted One 151's Weblog Says:

    […] the integrand is an even function of x, and so . Next, one can see that , and thus , and so . I previously proved that for a>0, , and therefore , for a>0, and so, for all real a, we have , as with our other […]

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