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I am trying the derive a path integral representation. I understand this involves Gaussian integrals of the form: $$\int_{-\infty}^\infty e^{-x^2}\text dx=\sqrt\pi$$ However, I am trying to evaluate the same integral but where $$\frac{i\, \delta t\, p^2}{2m}=x^2.$$ How do you substitute this into the original Gaussian? I tried integration by parts but it didn't work.

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marked as duplicate by Dimensio1n0, Qmechanic Mar 3 at 12:49

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Can you show us the exact integral you're having problems with? Might be I don't understand path integral very well, but from the information you provided, I see no reason why the result of the integral would be different. –  namehere Mar 3 at 10:23
    
I just have no idea where to start. –  user34039 Mar 3 at 10:27
    
What is your integration variable? It would be really helpful, as namehere said, to show us the exact integral you are trying to evaluate. –  Hunter Mar 3 at 10:35
    
What do you mean? It's the substitution for x –  user34039 Mar 3 at 10:36
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@user34039 I'm sorry but it appears your mathematical background is simply insufficient to properly approach this topic. Substitutions in integrals should be mastered by taking a course in calculus before attempting to tackle this theory (which requires considerably more difficult mathematical techniques like complex contour integrals to understand it properly). –  Danu Mar 3 at 11:15
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2 Answers 2

You need a slightly more general form for the integral, and in particular you need a form that allows for a constant in front of the $x^2$ in the exponent. To do this, using the substitution $y=\sqrt{a}x$, you do $$ \int_{-\infty}^\infty e^{-a x^2}\text dx= \int_{-\infty}^\infty e^{-y^2}\frac{\text dy}{\sqrt{a}}= \sqrt{\frac{\pi}{a}}. $$ You can then do your substitution, changing $x$ for $p$ and choosing an appropriate value of $a$.

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Can you please answer in terms of my specific substitution. –  user34039 Mar 3 at 10:44
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No. I will not do your work for you. I can point out the way to answer it, but that's as far as I will go. If you need clearer answers you should start with clearer questions. –  Emilio Pisanty Mar 3 at 10:45
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You asked the exact same thing here and seemed satified with the answers and comments. I suggest you have a look at integration by substitution. But to give you peace of mind, let's spell it out. Assuming $a\in\mathbb{C}, a \neq 0$ and setting $y^2=ax^2$ $$ \int_{-\infty}^{+\infty}e^{-ax^2}dx = \int_{-\infty}^{+\infty}e^{-y^2}\frac{dx}{dy}dy=\sqrt{a}^{-1}\int_{-\infty}^{+\infty}e^{-y^2}dy =\sqrt{\frac{\pi}{a}}$$

Edit: As Emilio Pisanty pointed out, for purely imaginary $a$, the integral is know as the Fresnel Integral and contrary to my former remark does indeed converge. The value can be conviniently calculated by means of complex calculus. But at the end of the day this procedure justifies the substitution in the Gaussian integral.

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So how did Zee do it? –  user34039 Mar 3 at 11:06
    
As it happens, for imaginary $a$ the integral is a [Fresnel integral]() that still converges (though it is not absolutely convergent). –  Emilio Pisanty Mar 3 at 11:15
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