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 $f$.

  • $\begingroup$ 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

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

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


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}


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 $ax \mapsto x$

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


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