4
$\begingroup$

While solving this trigonometric integral : $$\int_0^\infty \sin\left(a^2x^2+\frac{b^2}{x^2}\right)\mathrm{d}x$$ I came across this formula : $$\int_0^\infty f\left( a^2x^2+\frac{b^2}{x^2}\right)\mathrm{d}x=\frac{1}{a}\int_0^\infty f(x^2+2ab)\mathrm{d}x$$ And I'm wondering if I can prove it, but I hadn't any idea, because it seems like it depends on the properties of the function $f$.

$\endgroup$
4
  • $\begingroup$ Is there a term missing, or an extra + sign? $\endgroup$
    – Alex Jones
    Apr 1 at 20:21
  • 1
    $\begingroup$ Oh I'm sorry, it's just an extra (+), thanks for your remark ! $\endgroup$
    – Med-Elf
    Apr 1 at 20:22
  • $\begingroup$ I don't think that's true. How did you come across that formula? $\endgroup$
    – leonbloy
    Apr 1 at 20:29
  • 2
    $\begingroup$ Hi, so $$a^2x^2+\frac{b^2}{x^2}=a^2x^2-2ab+\frac{b^2}{x^2}+2ab = \left(ax-\frac bx\right)^2+2ab.$$ The term inside the function has a minimum. It will traverse from $+\infty$ to $2ab$ and then back to $+\infty$. In the other integral the value only traverses from $2ab$ to $+\infty$. However, making the substitutions introduces other terms that are missing in your formula. $\endgroup$
    – Snake707
    Apr 1 at 20:32
4
$\begingroup$

Assume that $a,b > 0$. If $\displaystyle{t = a x - \frac{b}{x}}$, then

$$t^2 + 2 a b = a^2 x^2 + \frac{b^2}{x^2} - 2 ab + 2 a b = a^2 x^2 + \frac{b^2}{x^2}.$$

Also one has

$$dt = \left(a + \frac{b}{x^2} \right) dx$$

as $x$ varies from $0$ to $\infty$, we see that $t$ varies from $-\infty$ to $\infty$. Hence, as long as one is careful about convergence, you formally get

$$\int_{-\infty}^{\infty} f(t^2 + 2 a b) dt = \int_{0}^{\infty} f \left(a^2 x^2 + \frac{b^2}{x^2} \right) \left(a + \frac{b}{x^2} \right) dx$$ $$ = a \int_{0}^{\infty} f \left(a^2 x^2 + \frac{b^2}{x^2} \right) dx + \int_{0}^{\infty} f \left(a^2 x^2 + \frac{b^2}{x^2} \right) \frac{b \cdot dx}{x^2}.$$

In the second integral, make the substitution $x \mapsto (b/a) u^{-1}$. This replaces $a^2 x^2$ by $b^2/u^2$ and $b^2/x^2$ to $a^2 u^2$, it changes the integrand from $[0,\infty)$ to $(\infty,0]$, and it changes

$$\frac{b \cdot dx}{x^2} = \frac{b \cdot d((b/a) u^{-1})}{(b/a)^2 u^{-2}} = \frac{- (b^2/a) u^{-2}}{b^2 a^{-2} u^{-2}} = -a \cdot du,$$

and thus (with the minus sign accounting for reversing the integrand) $$\int_{-\infty}^{\infty} f(t^2 + 2 a b) dt = a \int_{0}^{\infty} f \left(a^2 x^2 + \frac{b^2}{x^2} \right) dx + a \int_{0}^{\infty} f \left(a^2 u^2 + \frac{b^2}{u^2} \right) du.$$

Since the LHS is symmetric it $t \mapsto -t$, replacing the integrand by $[0,\infty)$ divides both sides by $2$ giving

$$\frac{1}{a} \int_{0}^{\infty} f(t^2 + 2 a b) dt = \int_{0}^{\infty} f \left(a^2 x^2 + \frac{b^2}{x^2} \right) dx.$$

$\endgroup$
2
$\begingroup$

With $a,\>b>0$

\begin{align} \int_0^\infty f\left( a^2x^2+\frac{b^2}{x^2}\right){d}x \overset{x\to \frac b{ax}}= & \int_0^\infty f\left( a^2x^2+\frac{b^2}{x^2}\right) \frac b{ax^2} dx\\ = & \frac12 \int_0^\infty f\left( a^2x^2+\frac{b^2}{x^2}\right) \left(1+\frac b{ax^2}\right) dx\\ = & \frac1{2a}\int_0^\infty f\left( (ax-\frac bx)^2+2ab\right) d\left(ax-\frac b{x}\right)\\ =& \frac1{2a}\int_{-\infty}^\infty f(x^2+2ab)dx\\ =&\frac1{a}\int_{0}^\infty f(x^2+2ab)dx \end{align}

$\endgroup$
1
$\begingroup$

Extend the integral using its evenness and complete the square on the inside

$$I = \frac{1}{2}\int_{-\infty}^{\infty}f\left(a^2x^2+\frac{b^2}{x^2}\right)dx = \frac{1}{2}\int_{-\infty}^{\infty}f\left(a^2\left[x - \frac{b}{ax}\right]^2 + 2ab\right)dx$$

Then use the following theorem (in this form attributed to Glasser's Master Theorem, but also known to Cauchy)

$$\int_{-\infty}^\infty g\left(x-\frac{k}{x}\right)dx = \int_{-\infty}^\infty g(x)\:dx$$

for $k>0$. With this, the final integral is

$$\frac{1}{2}\int_{-\infty}^{\infty}f\left(a^2\left[x - \frac{b}{ax}\right]^2 + 2ab\right)dx = \frac{1}{2}\int_{-\infty}^{\infty}f\left(a^2x^2 + 2ab\right)dx$$

which we can use the even property on again and use the variable exchange $ax \mapsto x$

$$\frac{1}{2}\int_{-\infty}^{\infty}f\left(a^2x^2 + 2ab\right)dx = \frac{1}{a}\int_{0}^{\infty}f\left(x^2 + 2ab\right)dx$$

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.