Recent content by rocdoc

Might a full proof of (8.18) involve using the following primitive, which is given as result 7.4.32 of Abramowitz and Stegun, see pg.303 of Reference 2? I quote '$$\int e^{-(ax^2+2bx+c)}\:\mathrm{d}x=\frac{1} {2}\sqrt{\frac{\pi} {a} }e^{ \frac{b^2-ac}{a}} erf(\sqrt ax+\frac{b} {\sqrt a})+const...

Readers of the following thread might be interested https://www.physicsforums.com/threads/decomposing-a-certain-exponential-integral.944617/ Also, those who are interested in path integrals in quantum theory and (8.18) in particular, might be interested in the material.

For the overall proof of (8.18), see the following thread https://www.physicsforums.com/threads/a-proof-of-kaku-8-18.945100/

In the following there is a proof, for positive values of ##a## only, of (8.18) of Kaku, reference 1, I quote' $$\int_{-\infty}^\infty~\mathrm{d}p~e^{iap^2+ibp}=\sqrt \frac{i\pi}{a}e^{-ib^2/4a}~~~~~~~~~~~~~(8.18)$$ '. Kaku says this result can be proved by completing the square. $$iap^2+ibp =...

A related thread, for those interested in path integrals in quantum theory/ quantum field theory is https://www.physicsforums.com/threads/path-integrals-in-quantum-theory.944350/

Perhaps even EQ(7) is correct?

It's looking good now.

One thing I should change is, I should put $$\int_{-\infty}^\infty~e^{i[a(p+\alpha)^2+\beta]}\mathrm{d}p=e^{i\beta}\int_{-\infty}^\infty~e^{i[a(p+\alpha)^2]}\mathrm{d}p$$

To the result 7.1.16 of Abramowitz and Stegun, Reference 1 of post1 ,one may add $$~~~~~~~~~~~~~~~erf~ z \rightarrow -1~(~z\rightarrow -\infty~~in~~~~|~\pi - arg~z|<\frac{\pi} {4})~~~~~~(2)$$ by use of symmetry, i.e. $$erf(-z)=-erf~~z~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~(3)$$ So I guess EQ(1) should be...

I have started to try to get into path integral formalism in quantum field theory. One thing I have tried to do is to prove a result from Kaku, reference 2, in the way that the author seems to suggest,from Kaku I quote' $$\int_{-\infty}^\infty~\mathrm{d}p~e^{iap^2+ibp}=\sqrt...

I have been trying to use, with##~a, \alpha ~\text{and}~ \beta## all real , and ##a## positive, the following $$\int_{-\infty}^\infty~e^{i[a(p+\alpha)^2+\beta]}\mathrm{d}p=\int_{-\infty}^\infty~\cos[a(p+\alpha)^2+\beta]\mathrm{d}p+i\int_{-\infty}^\infty~\sin[a(p+\alpha)^2+\beta]\mathrm{d}p$$ $$...

I don't suppose $$\tan \beta = \frac{(\cos\beta + \sin\beta) } {(\cos\beta - \sin\beta) }$$ It would be really nice if it were.

I show here , details of working out the expressions for the integrals that I gave previously. Let $$I_{cos}=\int_{-\infty}^\infty~\cos[a(p+\alpha)^2+\beta]\mathrm{d}p~~~~~~~~~~~~(1)$$ $$I_{sin}=\int_{-\infty}^\infty~\sin[a(p+\alpha)^2+\beta]\mathrm{d}p~~~~~~~~~~~~(2)$$ Use in EQ(1)...

In my case with ##~a, \alpha ~\text{and}~ \beta## all real, and with ##a## positive, I find , $$\int_{-\infty}^\infty~\cos[a(p+\alpha)^2+\beta]\mathrm{d}p=\sqrt \frac{ \pi} {2a} ( \cos\beta-\sin\beta )$$ $$\int_{-\infty}^\infty~\sin[a(p+\alpha)^2+\beta]\mathrm{d}p=\sqrt \frac{ \pi} {2a} (...

There is nothing wrong with the well known $$e^{i\theta}=\cos\theta+i\sin\theta$$ for real ## \theta## but what about $$\int_{-\infty}^\infty~e^{i\theta(p)}\mathrm{d}p=\int_{-\infty}^\infty~\cos\theta(p)\mathrm{d}p+i\int_{-\infty}^\infty~\sin\theta(p)\mathrm{d}p$$ I have been trying to use...