Prove that \,dx=\int_{-\infty}^{\infty}f(x)\,dx)
(with a>0).
First, we break up the improper integral at the x=0 singularity:
\,dx&=&\int_{-\infty}^{0^{-}}f\left(x-\frac{a}{x}\right)\,dx+\int_{0^{+}}^{\infty}f\left(x-\frac{a}{x}\right)\,dx\\&=&\lim_{b\to0^{-}}\int_{-\infty}^{b}f\left(x-\frac{a}{x}\right)\,dx+\lim_{c\to0^{+}}\int_{c}^{\infty}f\left(x-\frac{a}{x}\right)\,dx\end{eqnarray})
.
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
\,du)
;
and we see 
as 
, and 
as 
, and so, making the substitution,
\,dx=\int_{-\infty}^{\infty}f(u)\left(\frac{1}{2}-\frac{u}{2\sqrt{u^2+4a}}\right)\,du)
.
Similarly, for our second integral, x>0, and we have
,
so
\,du)
;
and we see 
as , and 
as , and so, making the substitution,
\,dx=\int_{-\infty}^{\infty}f(u)\left(\frac{1}{2}+\frac{u}{2\sqrt{u^2+4a}}\right)\,du)
.
Combining these,
\,dx&=&\int_{-\infty}^{0^{-}}f\left(x-\frac{a}{x}\right)\,dx+\int_{0^{+}}^{\infty}f\left(x-\frac{a}{x}\right)\,dx\\&=&\int_{-\infty}^{\infty}f(u)\left(\frac{1}{2}-\frac{u}{2\sqrt{u^2+4a}}\right)\,du+\int_{-\infty}^{\infty}f(u)\left(\frac{1}{2}+\frac{u}{2\sqrt{u^2+4a}}\right)\,du\\&=&\int_{-\infty}^{\infty}f(u)\left(\frac{1}{2}-\frac{u}{2\sqrt{u^2+4a}}+\frac{1}{2}+\frac{u}{2\sqrt{u^2+4a}}\right)\,du\\&=&\int_{-\infty}^{\infty}f(u)\,du\end{eqnarray})
.
Q.E.D.
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Tags: Integration, Math, Monday Math
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June 7, 2010 at 12:12 am |
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