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I am learning how to use Feynman's trick to solve integartion. I want to solve well known $\int _ {-\infty}^{\infty} e^{-x^2}dx$. I have seen the question Integrating $\int^{\infty}_0 e^{-x^2}\,dx$ using Feynman's parametrization trick but found nothing like mine.

I my attempt:

$$I(\alpha):= \int _ {-\infty}^{\infty} e^{-(x + \alpha)^2}dx\\ I'(\alpha)=\int _ {-\infty}^{\infty} e^{-(x + \alpha)^2} 2(-x-\alpha) dx\\ =-\int _ {x=-\infty}^{x=\infty} 2te^{-t^2}dt \quad [\text{where, } t = (x+\alpha)]\\ =e^{-t^2} \Biggr|_{x=-\infty}^{x=\infty}=e^{-(x+\alpha)^2}\Biggr|_{x=-\infty}^{x=\infty}=0$$

Hence, $$I(\alpha) = c \,\,\,\,(\text{constant})$$

Now, $$\lim_{\alpha \to \infty}I(\alpha) = 0 \implies c =0$$

Where I did mistake or we can't take $I(\alpha)$ like that please help. Thanks.

O M
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    Your integral $I(\alpha)$ is actually independent from $\alpha$ (i.e. your $I(\alpha)$ is a constant) because a substitution $u=x+\alpha$ will not change the bounds and $du=dx$. The reason you get $c=0$ is because $\displaystyle\lim_{\alpha\to+\infty}I(\alpha)=0$ is not true. – Angae MT Sep 02 '24 at 14:21

1 Answers1

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Well making a substitution of $u = x + \alpha$ we have: $$I(\alpha) = \int_{-\infty}^\infty e^{-(x + \alpha)^2} dx = \int_{-\infty}^\infty e^{-u^2} du = I(0)$$ So $I$ is a constant function, and so has derivative zero. It isn't true that $I(\alpha) \to 0$ as $\alpha \to \infty$. It is true that $e^{-(x + \alpha)^2} \to 0$ as $\alpha \to \infty$ for each fixed $x$, but this is not sufficient to conclude that $\int_{-\infty}^\infty e^{-(x + \alpha)^2} dx \to 0$. Note that $e^{-(x + \alpha)^2}$ is close to $1$ when $x$ is close to $-\alpha$ and think about what the transformation $x \mapsto x + \alpha$ is actually doing (the curve's shape and the area under it remains the same, it's just being shifted to the left, so it doesn't make sense that the integral will go to $0$ as you push $\alpha \to \infty$).

George Coote
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    this might be a bit advanced for you at the moment but you might want to peek at https://en.wikipedia.org/wiki/Dominated_Convergence_Theorem and https://math.libretexts.org/Bookshelves/Analysis/Real_Analysis_(Boman_and_Rogers)/08%3A_Back_to_Power_Series/8.02%3A_Uniform_Convergence-_Integrals_and_Derivatives for some examples of when you can move limits into integrals. In particular we do not have $e^{-(x + \alpha)^2} \to 0$ uniformly since, as above, given any $\alpha$, there are $x$ for which $e^{-(x + \alpha)^2}$ is $1$ or close to it. – George Coote Sep 02 '24 at 14:29
  • Thanks . "(the curve's shape and the area under it remains the same, it's just being shifted to the left, so it doesn't make sense that the integral will go to 0 as you push α→∞ )" was enough for me , it is nice explanation. I have a question: Is it means that we should use parameter multiplied with the $x$ which change the shape of the curve in feynman's trick? – O M Sep 02 '24 at 14:38
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    In this case that will require computing $\int_{-\infty}^\infty x^2 e^{-x^2} dx$ which is just as hard. You would usually calculate it by using IBP, where you'll need to know how to compute $\int_{-\infty}^\infty e^{-x^2} dx$. I'm sure there's techniques that don't need you to directly calculate the latter integral, maybe another DUTIS or some contour integral stuff – George Coote Sep 02 '24 at 14:46