Decomposing a Certain Exponential Integral

Click For Summary

Discussion Overview

The discussion revolves around the evaluation of certain exponential integrals, particularly focusing on the decomposition of integrals involving complex exponentials into their real and imaginary components. Participants explore the convergence of these integrals and their implications in the context of quantum field theory.

Discussion Character

  • Technical explanation
  • Mathematical reasoning
  • Debate/contested

Main Points Raised

  • One participant questions the validity of decomposing the integral of an exponential function into cosine and sine components, suggesting that the integrals may not converge.
  • Another participant provides specific results for the integrals of cosine and sine functions, asserting that they converge under certain conditions involving real parameters.
  • Details are shared on the derivation of these integrals, referencing a mathematical handbook for support.
  • A participant expresses skepticism about the integrals' convergence but acknowledges the correctness of the derivation presented.
  • There is a proposal to relate the derived results to a known equation from quantum field theory, raising concerns about discrepancies between the two approaches.
  • Further exploration of the relationship between the derived results and the reference from Kaku is discussed, with a suggestion to modify an equation for clarity.
  • Participants express tentative agreement on the validity of the derived expressions but maintain an open stance regarding their correctness.

Areas of Agreement / Disagreement

Participants do not reach a consensus on the convergence of the integrals or the correctness of the derived expressions. Multiple competing views remain regarding the implications of the results in the context of quantum field theory.

Contextual Notes

Some integrals are noted to be infinite, raising questions about their convergence. The discussion includes references to specific mathematical results and assumptions that may not be universally accepted.

Who May Find This Useful

This discussion may be of interest to those studying mathematical methods in physics, particularly in relation to quantum field theory and the evaluation of complex integrals.

rocdoc
Gold Member
Messages
43
Reaction score
3
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?
 
Physics news on Phys.org
The integrals that you show usually will not be convergent.
 
mathman said:
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 ,
$$\int_{-\infty}^\infty~\cos[a(p+\alpha)^2+\beta]\mathrm{d}p=\sqrt \frac{ \pi} {2a} ( \cos\beta-\sin\beta )$$
$$\int_{-\infty}^\infty~\sin[a(p+\alpha)^2+\beta]\mathrm{d}p=\sqrt \frac{ \pi} {2a} ( \cos\beta+\sin\beta )$$
 
Last edited:
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.
 
I don't suppose
$$\tan \beta = \frac{(\cos\beta + \sin\beta) } {(\cos\beta - \sin\beta) }$$
It would be really nice if it were.
 
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
$$I_{Tot}=\sqrt { \frac{ \pi} {a}} \exp (i~tan^{-1}(\frac{ \cos\beta + \sin\beta }{ \cos\beta - \sin\beta })~~)$$
 
The integral of the absolute value of the integrands are infinite, so I was a little skeptical. However, your derivation seems to be correct.
 
  • Like
Likes   Reactions: rocdoc
jedishrfu said:
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.
 
  • #10
One thing I should change is, I should put

$$\int_{-\infty}^\infty~e^{i[a(p+\alpha)^2+\beta]}\mathrm{d}p=e^{i\beta}\int_{-\infty}^\infty~e^{i[a(p+\alpha)^2]}\mathrm{d}p$$
 
  • #11
It's looking good now.
 
  • #12
Perhaps even EQ(7) is correct?
 

Similar threads

Undergrad Inequality with integral and max of derivative
  • · Replies 3 ·
Replies
3
Views
3K
Undergrad Solving Stat Mech Integral with Wolfram Alpha
  • · Replies 2 ·
Replies
2
Views
2K
Undergrad Why does the integral of sine of x^2 from - infinity to + infinity diverge?
  • · Replies 4 ·
Replies
4
Views
3K
Undergrad Having trouble understanding a seemingly simple u-substitution
  • · Replies 5 ·
Replies
5
Views
2K
Graduate Challenging integral involving exponentials and logarithms
  • · Replies 14 ·
Replies
14
Views
3K
High School Having trouble deriving the volume of an elliptical ring toroid
  • · Replies 4 ·
Replies
4
Views
4K
Graduate Summing over continuum and uncountable numerocities
  • · Replies 1 ·
Replies
1
Views
3K
Undergrad On convolution theorem of Laplace transform: Schiff
  • · Replies 4 ·
Replies
4
Views
3K
Graduate Solving an Integral involving a probability density function
  • · Replies 16 ·
Replies
16
Views
3K
Undergrad Integration Using Trigonometric Substitution
  • · Replies 8 ·
Replies
8
Views
3K