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The problem says,

If $\int_{0}^{\infty} \mathbb{e^{-(a^2x^2+\frac{b^2}{x^2})dx}=\frac{\sqrt{\pi}}{2a}.e^{-2ab}} \longrightarrow(i)$, then prove that $\mathbb{\int_{0}^{\infty}\frac{1}{x^2}e^{-(a^2x^2+\frac{b^2}{x^2})}dx=\frac{\sqrt{\pi}}{2b}.e^{-2ab}}$

If we take $\mathrm{x=\frac{b}{a\mathscr{X}}}$, where $\mathrm{a}, \mathrm{b}$ are any constant, then we have $$\mathbb{\int_{0}^{\infty}\frac{1}{x^2}e^{-(a^2x^2+\frac{b^2}{x^2})}dx} \longrightarrow (ii)\\=\mathbb{e^{-2ab}}\mathbb{\int_{0}^{\infty}\frac{1}{x^2}e^{-(ax+\frac{b}{x})^2}dx}\\=\mathbb{\frac{-a\ e^{-2ab}}{b}}\mathbb{\int_{\infty}^{0}e^{-(b\mathscr{X}+\frac{a}{\mathscr{X}})^2}d\mathscr{X}}\\=\mathbb{\frac{a}{b}}\mathbb{\int_{0}^{\infty}e^{-(b^{2}\mathscr{X}^2+\frac{a^2}{\mathscr{X}^2})}d\mathscr{X}}\longrightarrow(iii)\\$$

  • Since $\mathrm{a}, \mathrm{b}$ are any constant and condition $(i) \ \textrm{&} \ (iii)$ look same and the integration range remains same, therefore should we use the value of $(i)$ in $(iii)$ ? If we plug in the value of $(i)$ in $(iii)$, the desired result comes. But is this a correct aprroach to do that?

Any help, explanation is valuable and highly appreciated.

vbm
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    The second integral in the question you've written in "The problem says" contains a factor of $\frac{1}{x^2}$ in the integrand, which isn't present in the first integral. The title of this question ignores that. Is that intentional? – John Hughes Jul 03 '20 at 11:26
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    If the $\frac{1}{x^2}$ is indeed in the integral, then all you have to do is apply the substitution $t = \frac{1}{x}$ and then use the first formula (the roles of $a$ and $b$ will have switched), so that approach is fine. – Ninad Munshi Jul 03 '20 at 11:29
  • @JohnHughes, I was just thinking if I can use the value in condition i in condition iii. So I just mentioned the integrals only.. which are from $0$ to $\infty$ – vbm Jul 03 '20 at 11:34
  • @NinadMunshi I wanted to avoid using $1$ as you mentioned $t=\frac{1}{x}$ and I was thinking if I can solve the problem for any two arbitrary constants like $a,b$ – vbm Jul 03 '20 at 11:37

1 Answers1

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You can "plug i into iii" via the "integration-by-substitution" $$ \mathscr{X} = x \\ d \mathscr{X} = dx $$ (i.e., using the chain rule in the simplest possible way); this shows that the integral you're calling "iii" is the same as the integral you're calling "i".

John Hughes
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  • +1 for the clear answer. Following your answer may I say that $a,b$ are "arbitrary", so swapping the places of $a$ and $b$ in any of the integrand does not change the value significantly? – vbm Jul 03 '20 at 11:58
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    No, you cannot say that. But you can replace the symbols $a$ and $b$ by, say, $B$ and $A$ to get a new and equally valid formula, and then you can replace $B$ and $A$ by $b$ and $a$ similarly. As an example, $\int \frac{a}{x+b} = a \log(x+b)$; you're not allowed to say "a and b are arbitrary, so swapping them won't change the value", because that would produce $\int \frac{b}{x + a} = a \log (x + b)$, which is generally false. In short: you can rename variables in an equality, but you must do so everywhere they appear, not just inside the integrand. – John Hughes Jul 03 '20 at 12:07
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    I mean that that saying "integral of this equals that" is logically derivable from "integral of THIS equals THAT" (and vice versa) in a particularly simple way, namely, replacement of variable-names throughout a formula. It's just like saying that "3x + x = 4x" is "the same as" the formula "3a + a = 4a", by swapping the variable "x" for the variable "a". They're not character-for-character the same, but they are so very slightly different that it's not worth writing both down. Implicitly, these are quantified formulas ($\forall x, 3x + x = 4x$), and I'm replacing the bound variable. – John Hughes Jul 03 '20 at 12:33
  • Alright, it is clear now to me and thank you for your clear answer and helpful comments. – vbm Jul 03 '20 at 12:35