$$ \int\limits_0^{+\infty} e^{-a^2x^2-\frac{b^2}{x^2}} dx$$
I'm stuck after doing this:
$$ I = e^{-2ab}\int\limits_0^{+\infty} e^{-(ax-\frac{b}{x})^2} dx$$
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After doing that, set $x=z\sqrt{\frac{b}{a}},\;dx = dz\sqrt{\frac{b}{a}}$ and apply Glasser's master theorem or just the
Lemma. If $f\in L^1(\mathbb{R})$, $$ \int_{-\infty}^{+\infty}f(x)\,dx = \int_{-\infty}^{+\infty}f\left(x-\frac{1}{x}\right)\,dx.$$
Jack D'Aurizio
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1And then use Gaussian integral to calculate: $$ I = \frac{\sqrt{\pi}}{2a},e^{-2ab} $$ – Hazem Orabi Mar 01 '17 at 11:22
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How to prove this lemma? – Mannix Mar 01 '17 at 14:51
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@Mannix: by spliiting the integration range as $(0,1)\cup(1,+\infty)$ and applying the substitution $x-\frac{1}{x}=z$ in both the resulting integrals, leading to an unexpected cancellation. – Jack D'Aurizio Mar 01 '17 at 15:01
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By that lemma, $$ \int\limits_0^{\infty} \frac{1}{x^2}dx = \int\limits_0^{\infty} \frac{1}{(x-\frac{1}{x})^2} dx $$, isn't it? – Mannix Mar 01 '17 at 20:01
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@Mannix: no, since $\frac{1}{x^2}$ is not integrable over $\mathbb{R}^+$. – Jack D'Aurizio Mar 01 '17 at 20:28
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Why? It's intergal is finite – Mannix Mar 01 '17 at 20:39
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@Mannix: really? And what is the value of $\int_{0}^{+\infty}\frac{dx}{x^2}$, if it is finite? – Jack D'Aurizio Mar 01 '17 at 20:40
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I'm sorry, it really doesn't converge, but for $f(x)=\frac{1}{(x+1)^2}$, lemma must work, while $\int\limits_0^{\infty} \frac{1}{(x-1/x+1)^2} $ doesn't converge – Mannix Mar 01 '17 at 20:48
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@Mannix: oh, sorry, I mentioned GMT on $\mathbb{R}^+$, but the integration range should have been $\mathbb{R}$. Now fixed. – Jack D'Aurizio Mar 01 '17 at 20:54