Integration by Inversion

Theorem

$\ds \int_0^{+\infty} \map f x \rd x = \int_0^{+\infty} \dfrac {\map f {\frac 1 x} } {x^2} \rd x$

Proof

\(\ds \int_0^{+\infty} \map f x \rd x\)

\(=\)

\(\ds \int_{x \mathop \to 0}^{x \mathop \to +\infty} \map f x \rd x\)

Definition of Improper Integral

\(\ds \)

\(=\)

\(\ds \int_{\frac 1 x \mathop \to 0}^{\frac 1 x \mathop \to +\infty} \map f {\frac 1 x} \map \rd {\frac 1 x}\)

Integration by Substitution

\(\ds \)

\(=\)

\(\ds \int_{\frac 1 x \mathop \to 0}^{\frac 1 x \mathop \to +\infty} \map f {\frac 1 x} \paren {-\dfrac 1 {x^2} } \rd x\)

Power Rule for Derivatives

\(\ds \)

\(=\)

\(\ds -\int_{x \mathop \to +\infty}^{x \mathop \to 0} \dfrac {\map f {\frac 1 x} } {x^2} \rd x\)

simplifying

\(\ds \)

\(=\)

\(\ds \int_{x \mathop \to 0}^{x \mathop \to +\infty} \dfrac {\map f {\frac 1 x} } {x^2} \rd x\)

Reversal of Limits of Definite Integral

\(\ds \)

\(=\)

\(\ds \int_0^{+\infty} \dfrac {\map f {\frac 1 x} } {x^2} \rd x\)

Definition of Improper Integral

$\blacksquare$