Decomposing a Certain Exponential Integral

[rocdoc]

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, with$~a, \alpha ~and~ \beta$ all real

$$\int_{-\infty}^\infty~e^{i[a(p+\alpha)^2+\beta]}\mathrm{d}p=\int_{-\infty}^\infty~\cos[a(p+\alpha)^2+\beta]\mathrm{d}p+i\int_{-\infty}^\infty~\sin[a(p+\alpha)^2+\beta]\mathrm{d}p$$
is this OK?

 

[jedishrfu]

It seems reasonable:

https://en.wikipedia.org/wiki/Integration_using_Euler's_formula

Have you run into a problem?

 

[mathman]

The integrals that you show usually will not be convergent.

 

[rocdoc]

The integrals that you show usually will not be convergent.

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

 

[rocdoc]

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)
$$\cos(~A+B~)=\cos A\cos B-\sin A\sin B $$
$$I_{cos}=\int_{-\infty}^\infty~\{\cos[a(p+\alpha)^2]\cos\beta-\sin[a(p+\alpha)^2]\sin\beta\}\mathrm{d}p~~~~~~~~~~~~$$
$$I_{cos}=\cos\beta~\int_{-\infty}^\infty~\\cos[a(p+\alpha)^2]\mathrm{d}p - \sin\beta~\int_{-\infty}^\infty~\sin[a(p+\alpha)^2]\mathrm{d}p $$
Substitute $q=p+\alpha$ giving
$$\int_{-\infty}^\infty~\\cos[a(p+\alpha)^2]\mathrm{d}p=\int_{-\infty}^\infty~\\cos(aq^2)\mathrm{d}q$$
$$\int_{-\infty}^\infty~\\sin[a(p+\alpha)^2]\mathrm{d}p=\int_{-\infty}^\infty~\\sin(aq^2)\mathrm{d}q$$
$$I_{cos}=\cos\beta~\int_{-\infty}^\infty~\\cos(aq^2)\mathrm{d}q - \sin\beta~ \int_{-\infty}^\infty~\\sin(aq^2)\mathrm{d}q$$
Use Spiegel, see Reference 1 result 15.50, which I quote '
$$15.50~~~~~~~~\int_0^\infty~\\cos(ax^2)\mathrm{d}x = \int_0^\infty~\\sin(ax^2)\mathrm{d}x= \frac{ 1} {2} \sqrt \frac{ \pi} {2a} $$
' ,for positive $a$ hence as the integrands in 15.50 are even functions of $x$
$$\int_{-\infty}^\infty~\\cos(ax^2)\mathrm{d}x = \int_{-\infty}^\infty~\\sin(ax^2)\mathrm{d}x= \sqrt \frac{ \pi} {2a}$$
So
$$I_{cos}=\cos\beta \sqrt \frac{ \pi} {2a} - \sin\beta \sqrt \frac{ \pi} {2a} $$
$$I_{cos}= \sqrt \frac{ \pi} {2a} (\cos\beta - \sin\beta ) $$
Similarly but using in EQ(2)
$$\sin(~A+B~)=\sin A\cos B+\cos A\sin B $$
$$I_{sin}= \sqrt \frac{ \pi} {2a} (\cos\beta + \sin\beta ) $$
Reference
1) M. R. Spiegel, Ph.D. , Mathematical Handbook , McGraw-Hill , Inc. , 1968.

 

[rocdoc]

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

 

[rocdoc]

I have been trying to use, with$~a, \alpha ~\text{and}~ \beta$ all real , and $a$ positive, the following
$$\int_{-\infty}^\infty~e^{i[a(p+\alpha)^2+\beta]}\mathrm{d}p=\int_{-\infty}^\infty~\cos[a(p+\alpha)^2+\beta]\mathrm{d}p+i\int_{-\infty}^\infty~\sin[a(p+\alpha)^2+\beta]\mathrm{d}p$$
$$ =I_{cos} + i~I_{sin}~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~$$
$$= \sqrt \frac{ \pi} {2a} (\cos\beta - \sin\beta ) +i \sqrt \frac{ \pi} {2a} (\cos\beta + \sin\beta ) $$
$$ =I_{Tot}~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~$$
So
$$ I_{Tot}=I_{cos}+i~I_{sin}~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~$$
$$= \sqrt \frac{ \pi} {2a} (\cos\beta - \sin\beta ) +i \sqrt \frac{ \pi} {2a} (\cos\beta + \sin\beta ) $$
I show here , some details of working out a polar form for $I_{Tot}$ .
For any complex number $z$
$$z=c+i~d~~; ~ z=re^{i\theta}~~~\text{where}~r=\sqrt{c^2+d^2},\theta=tan^{-1}\frac{d}{c}$$
Put
$$z=I_{Tot}~; c=\sqrt \frac{ \pi} {2a} (\cos\beta - \sin\beta ) ;~d= \sqrt \frac{ \pi} {2a} (\cos\beta + \sin\beta )$$
$$r=\sqrt{ \frac{ \pi} {a}}$$
$$\theta=tan^{-1}(\frac{ \cos\beta + \sin\beta }{ \cos\beta - \sin\beta })$$
Putting the above together
$$I_{Tot}=\sqrt { \frac{ \pi} {a}} \exp (i~tan^{-1}(\frac{ \cos\beta + \sin\beta }{ \cos\beta - \sin\beta })~~)$$

 

[mathman]

The integral of the absolute value of the integrands are infinite, so I was a little skeptical. However, your derivation seems to be correct.

 

[rocdoc]

Have you run into a problem?

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'
$$\int_{-\infty}^\infty~\mathrm{d}p~e^{iap^2+ibp}=\sqrt \frac{i\pi}{a}e^{-ib^2/4a}~~~~~~~~~~~~~(8.18)$$
'. Kaku says this result can be proved by completing the square.The way I have gone about this , is the following.
$$iap^2+ibp = ia(p^2+\frac{b}{a}p )$$
$$~~~~~~~~~~~~~~~~~~~~~~~~=ia[~(p+\frac{b}{2a})^2-\frac{b^2}{4a^2}]$$
Let
$$\frac{b}{2a}=\alpha~~~~~~~~~(3)$$
$$\frac{-b^2}{4a}=\beta~~~~~~~~~~(4)$$
$$iap^2+ibp = i[~a(p+\alpha)^2~+~\beta]$$
$$\int_{-\infty}^\infty~\mathrm{d}p~e^{iap^2+ibp}=\int_{-\infty}^\infty~e^{ i[~a(p+\alpha)^2~+~\beta] }\mathrm{d}p~~~~~~~(5)$$
The integrals in EQ(5) are what have been called $I_{Tot}$. According to EQ(8.18)
$$I_{Tot}=\sqrt \frac{i\pi}{a}e^{-ib^2/4a}~~~~~~~~$$
$$~~~~~~~~~~=\sqrt \frac{i\pi}{a}\exp(i\beta)~~~~~~~~~(6)$$
However , what has been derived in this thread is
$$I_{Tot}=\sqrt { \frac{ \pi} {a}} \exp (i~tan^{-1}(\frac{ \cos\beta + \sin\beta }{ \cos\beta - \sin\beta })~~)~~~~~~(7)$$

I have a problem with the above (for example).Reference:
2) M.Kaku , QuantumField Theory, A Modern Introduction , Oxford University Press, Inc. , 1993.

 

[rocdoc]

One thing I should change is, I should put

$$\int_{-\infty}^\infty~e^{i[a(p+\alpha)^2+\beta]}\mathrm{d}p=e^{i\beta}\int_{-\infty}^\infty~e^{i[a(p+\alpha)^2]}\mathrm{d}p$$

 

[rocdoc]

It's looking good now.

 

[rocdoc]

Perhaps even EQ(7) is correct?

 

[rocdoc]

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/

 

[rocdoc]

For the overall proof of (8.18), see the following thread

https://www.physicsforums.com/threads/a-proof-of-kaku-8-18.945100/