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rocdoc
Gold Member
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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.
$$iap^2+ibp = ia(p^2+\frac{b}{a}p )$$
$$~~~~~~~~~~~~~~~~~~~~~~~~=ia[~(p+\frac{b}{2a})^2-\frac{b^2}{4a^2}]$$
Let
$$\frac{b}{2a}=\alpha~~~~~~~~~(1)$$
$$\frac{-b^2}{4a}=\beta~~~~~~~~~~(2)$$
$$iap^2+ibp = i[~a(p+\alpha)^2~+~\beta]$$
Let
$$I=\int_{-\infty}^\infty~\mathrm{d}p~e^{iap^2+ibp}=\int_{-\infty}^\infty~e^{ i[~a(p+\alpha)^2~+~\beta] }\mathrm{d}p$$
$$~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~=e^{i\beta}~\int_{-\infty}^\infty~e^{ i[~a(p+\alpha)^2] }\mathrm{d}p~~~~~~~(3)$$
Concentrating upon the integral part of (3)
$$\int_{-\infty}^\infty~e^{ i[~a(p+\alpha)^2] }\mathrm{d}p=\int_{-\infty}^\infty \cos[a(p+\alpha)^2]\mathrm{d}p +i\int_{-\infty}^\infty \sin[a(p+\alpha)^2]\mathrm{d}p~~~~~~~~~~(4)$$
Substituting ##q=p+\alpha## gives
$$\int_{-\infty}^\infty~\cos[a(p+\alpha)^2]\mathrm{d}p=\int_{-\infty}^\infty~\cos(aq^2)\mathrm{d}q$$
$$\int_{-\infty}^\infty~\sin[a(p+\alpha)^2]\mathrm{d}p=\int_{-\infty}^\infty~\sin(aq^2)\mathrm{d}q$$
Use Spiegel, see Reference 2 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##
$$\int_{-\infty}^\infty~\cos(ax^2)\mathrm{d}x = \int_{-\infty}^\infty~\sin(ax^2)\mathrm{d}x= \sqrt \frac{ \pi} {2a}~~~~~~(5)$$
Using (5) in (4)
$$~~~~\int_{-\infty}^\infty~e^{ i[~a(p+\alpha)^2] }\mathrm{d}p= \sqrt \frac{ \pi} {2a} +i \sqrt \frac{ \pi} {2a}$$
In polar form
$$~\int_{-\infty}^\infty~e^{ i[~a(p+\alpha)^2] }\mathrm{d}p= \sqrt \frac{ \pi} {a} e^{i\pi/4}$$
Also, as
$$ ~~~~~~~~~~~ e^{i\pi/4}=\sqrt i$$
$$~~~\int_{-\infty}^\infty~e^{ i[~a(p+\alpha)^2] }\mathrm{d}p= \sqrt \frac{ i\pi } {a} ~~~~(6)$$
Hence, using (6) and (2) in (3)
$$I=e^{i\beta} \sqrt \frac{ i\pi } {a}=e^{-ib^2/4a} \sqrt \frac{i \pi} {a}$$
I.E.
$$~~~~~~~~~~~~~~~\int_{-\infty}^\infty~\mathrm{d}p~e^{iap^2+ibp}=\sqrt \frac{i\pi}{a}e^{-ib^2/4a}~~~~~~~~Q.E.D.$$References
1) M.Kaku , QuantumField Theory, A Modern Introduction , Oxford University Press, Inc. , 1993.
2) M. R. Spiegel, Ph.D. , Mathematical Handbook , McGraw-Hill , Inc. , 1968.
$$\int_{-\infty}^\infty~\mathrm{d}p~e^{iap^2+ibp}=\sqrt \frac{i\pi}{a}e^{-ib^2/4a}~~~~~~~~~~~~~(8.18)$$
'. Kaku says this result can be proved by completing the square.
$$iap^2+ibp = ia(p^2+\frac{b}{a}p )$$
$$~~~~~~~~~~~~~~~~~~~~~~~~=ia[~(p+\frac{b}{2a})^2-\frac{b^2}{4a^2}]$$
Let
$$\frac{b}{2a}=\alpha~~~~~~~~~(1)$$
$$\frac{-b^2}{4a}=\beta~~~~~~~~~~(2)$$
$$iap^2+ibp = i[~a(p+\alpha)^2~+~\beta]$$
Let
$$I=\int_{-\infty}^\infty~\mathrm{d}p~e^{iap^2+ibp}=\int_{-\infty}^\infty~e^{ i[~a(p+\alpha)^2~+~\beta] }\mathrm{d}p$$
$$~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~=e^{i\beta}~\int_{-\infty}^\infty~e^{ i[~a(p+\alpha)^2] }\mathrm{d}p~~~~~~~(3)$$
Concentrating upon the integral part of (3)
$$\int_{-\infty}^\infty~e^{ i[~a(p+\alpha)^2] }\mathrm{d}p=\int_{-\infty}^\infty \cos[a(p+\alpha)^2]\mathrm{d}p +i\int_{-\infty}^\infty \sin[a(p+\alpha)^2]\mathrm{d}p~~~~~~~~~~(4)$$
Substituting ##q=p+\alpha## gives
$$\int_{-\infty}^\infty~\cos[a(p+\alpha)^2]\mathrm{d}p=\int_{-\infty}^\infty~\cos(aq^2)\mathrm{d}q$$
$$\int_{-\infty}^\infty~\sin[a(p+\alpha)^2]\mathrm{d}p=\int_{-\infty}^\infty~\sin(aq^2)\mathrm{d}q$$
Use Spiegel, see Reference 2 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##
$$\int_{-\infty}^\infty~\cos(ax^2)\mathrm{d}x = \int_{-\infty}^\infty~\sin(ax^2)\mathrm{d}x= \sqrt \frac{ \pi} {2a}~~~~~~(5)$$
Using (5) in (4)
$$~~~~\int_{-\infty}^\infty~e^{ i[~a(p+\alpha)^2] }\mathrm{d}p= \sqrt \frac{ \pi} {2a} +i \sqrt \frac{ \pi} {2a}$$
In polar form
$$~\int_{-\infty}^\infty~e^{ i[~a(p+\alpha)^2] }\mathrm{d}p= \sqrt \frac{ \pi} {a} e^{i\pi/4}$$
Also, as
$$ ~~~~~~~~~~~ e^{i\pi/4}=\sqrt i$$
$$~~~\int_{-\infty}^\infty~e^{ i[~a(p+\alpha)^2] }\mathrm{d}p= \sqrt \frac{ i\pi } {a} ~~~~(6)$$
Hence, using (6) and (2) in (3)
$$I=e^{i\beta} \sqrt \frac{ i\pi } {a}=e^{-ib^2/4a} \sqrt \frac{i \pi} {a}$$
I.E.
$$~~~~~~~~~~~~~~~\int_{-\infty}^\infty~\mathrm{d}p~e^{iap^2+ibp}=\sqrt \frac{i\pi}{a}e^{-ib^2/4a}~~~~~~~~Q.E.D.$$References
1) M.Kaku , QuantumField Theory, A Modern Introduction , Oxford University Press, Inc. , 1993.
2) M. R. Spiegel, Ph.D. , Mathematical Handbook , McGraw-Hill , Inc. , 1968.
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