Feynman integral 1

Feynman's Integral 8.59 p 170 of BD1 PhL 11.1.08 Comments All integrals on this page are from 0 to ∞. When I first saw this integral, I tried for about 2 hours and then gave up on deriving it. Here is how I kept starting off: 1/a1a2 =∫dz1∫dz2 1/(a1z1+ a2z2)2 δ(z1+ z2 - 1) =(1/a1a2)∫dz1∫dz2 1/(z1+ z2)2 δ(z1/a1 + z2/a2 - 1) // scale zi→ zi/ai =(1/a1a2) I(ai) I then took the approach of trying to show why the resulting integral I(ai) was independent of the ai. If I could show that, then I could just set all ai = 1 and do the integral directly (though it was unclear I could do the integral for many variables). I tried taking the derivative of the RHS but that led to a δ' sitting in there, which I tried undoing by a parts integration, but that just led to a mess. I then tried elevating the delta function into an exponential, and that also led nowhere. At that point, I decided I was "missing some simple fact" and, since Halloween, let's go to Marriott and look up the Feynman reference given on BD page 171. I found Marriott in a state of major renovation. The inner core is empty, all periodicals are in some robotic machine called the ARC. [ email sent out just now on that subject ] I got the 1949 Feynman article referenced on page 171 from the ARC, and the volume had lots of articles by Feynman Schwinger and others on the new QED stuff, but I did not see my integral in any of these articles. They use A in place of and A in place of a. The problem is doing the Feynman graph integrals with their various denominator factors from propagators, and that is where this integral comes into play. I was amazed at the amount of math detail in these old pages, thousands of symbols per page in densepack tiny print. This was Phys Rev, before it broke out into A,B,C,D. Having failed on this, I wandered over to QC680 .J38 and looked at Jauch and Rorlich "The Theory of Photons and Electrons, another BD reference. This was 1955 and then a 1972 second edition by the surviving Rorlich (second on shelf, first in ARC). In an appendix, my integral appeared along with an induction proof which I copied. Then just before leaving I looked at a QED + QCD book by Grozin, and this had the simple proof I was looking for, no induction needed. Proof of Simple Case As I suspected, I was indeed missing a very simple fact which is this, ∫dx xne-ax = n!/an+1   (1) an old friend. So let's consider the RHS of our equation above J = ∫dz1∫dz2 1/(a1z1+ a2z2)2 δ(z1+ z2 - 1) The key step is to use the above integral (1) with n=1 to "elevate" the denominator factor into an integral representation: 1/(a1z1+ a2z2)2 = ∫dx x exp(-[a1z1+ a2z2]x) = ∫dx x exp(-[a1z1]x) exp(-[a2z2]x) and you can see how this tends to "factorize" things, separating a1 from a2. We then have J = ∫dx x ∫dz1∫dz2 exp(-[a1z1]x) exp(-[a2z2]x) δ(z1+ z2 - 1) Now is the time to "scale" variables by doing zi' = zi x to clean up the variables in the exponents, and we then get J = ∫dx x ∫dz'1∫dz'2 (1/x)2 exp(-[a1z'1]) exp(-[a2z'2]) δ(z'1/x+ z'2/x - 1) = ∫dx x ∫dz'1∫dz'2 (1/x)2 exp(-[a1z'1]) exp(-[a2z'2]) δ(z'1+ z'2 - x)(x) // adjust δ = ∫dx ∫dz'1∫dz'2 exp(-[a1z'1]) exp(-[a2z'2]) δ(z'1+ z'2 - x) // cancel x's = ∫dx ∫dz1∫dz2 exp(-[a1z1]) exp(-[a2z2]) δ(z1+ z2 - x) // remove primes = ∫dz1∫dz2 exp(-[a1z1]) exp(-[a2z2]) ∫dx δ(z1+ z2 - x) // move ∫dx to the right = ∫dz1∫dz2 exp(-[a1z1]) exp(-[a2z2]) // do the x integral = ∫dz1 exp(-[a1z1]) ∫dz2 exp(-[a2z2]) // result is fully factored = 1/a1 * 1/a2 QED In the above, the scaling keeps z1 and z2 on an equal footing, unlike my initial scaling shown at the start of this writeup. So we can regard the above as showing the basic idea for doing this kind of integral. Let's now do the full integral shown by Grozin, of which the Feynman integral on top of BD page 170 is a simple case. I will go in the "forward direction" instead of the reverse direction done above. Start with as many factors as you want on the LHS: LHS = [n1! / (a1n1+1) ] [n2! / (a2n2+1) ] ..... [nk! / (aknk+1) ] Throw each one into its integral representation using (1) above: LHS = ( ∫dx1 x1n1e-a1*x1) ( ∫dx2 x2n2e-a2*x2)..... ( ∫dxk xknke-ak*xk) Combine all the exponentials together LHS = ∫dx1∫dx2....∫dxk x1n1x2n2... xknk e-a1*x1-a2*x2.... -ak*xk Now just insert 1 = ∫dx δ(x1 + x2 + .... – x) as part of the integrand above: LHS = ∫dx1∫dx2....∫dxk x1n1x2n2... xknk e-a1*x1-a2*x2.... -ak*xk ∫dx δ(x1 + x2 + .... – x) and then shift the dx integral all the way to the left LHS = ∫dx ∫dx1∫dx2....∫dxk x1n1x2n2... xknk e-a1*x1-a2*x2.... -ak*xk δ(x1 + x2 + .... – x) Now scale all integration xi variables so that xi = x zi where x that new integration variable which we know is a positive number: LHS = ∫dx ∫dz1∫dz2....∫dzk (x)k * z1n1z2n2... zknk (x)n1+n2+..nk * e-[a1*z1+a2*z2.... +ak*zk]x * δ(xz1 + xz2 + .... – x) Now remove x from the delta function δ(xz1 + xz2 + .... – x) = δ(z1 + z2 + .... – 1 )/x and combine powers of x to get LHS = ∫dx ∫dz1∫dz2....∫dzk * z1n1z2n2... zknk * e-[a1*z1+a2*z2.... +ak*zk]x * δ(z1 + z2 + .... –1 ) xk-1+n1+n2+...nk So next, move the δ to the left and the dx integral to the right to get LHS = ∫dz1∫dz2....∫dzk * z1n1z2n2... zknk δ(z1 + z2 + .... –1 ) * ∫dx e-[a1*z1+a2*z2.... +ak*zk]x * xk-1+n1+n2+...nk and we see that the x integral is now of our form (1) above. Let's define these temporary names k-1+n1+n2+...nk = N a1*z1+a2*z2.... +ak*zk = Q ∫dx e-[Q]x * xN = (N)! / QN+1 and our result is LHS = N! ∫dz1∫dz2....∫dzk * z1n1z2n2... zknk /[Q] N+1 δ(z1 + z2 + .... –1 ) So we have now shown that [n1! / (a1n1+1) ] [n2! / (a2n2+1) ] ..... [nk! / (aknk+1) ] = N! ∫dz1∫dz2....∫dzk * z1n1z2n1... zknk /[Q] N+1δ(z1 + z2 + .... –1 ) which we can rewrite as [1/ (a1n1+1) ] [1 / (a2n2+1) ] ..... [1/ (aknk+1) ] = N! / (n1! n2!.... nk!) * ∫dz1∫dz2....∫dzk * z1n1z2n2... zknk /[Q] N+1δ(z1 + z2 + .... –1 ) where N = k-1+n1+n2+...nk and Q = a1*z1+a2*z2.... +ak*zk. Finally, define some new integers mi = ni+1 for i = 1 to k. then N = k-1+n1+n2+...nk = m1+m2+...mk - 1 which removes term k from N. Then we have [1/ (a1m1) ] [1 / (a2m2) ] ..... [1/ (akmk) ] = (m1+m2+...mk - 1)! / { (m1-1)! (m2-1)! ... (m1-1)! } * ∫dz1∫dz2....∫dzk * z1m1-1z2m2-1... zkmk-1 /[Q]m1+m2+..+mk δ(z1 + z2 + .... –1 ) And things are more compact if we write (n-1)! = Γ(n) and result is then 1/(a1m1a2m2 ....akmk) = Γ(m1+ m2+..mk)/ (Γ(m1) Γ(m2).. Γ(mk)) ∫dz1∫dz2....∫dzk * z1m1-1z2m2-1... zkmk-1 /[ a1*z1+a2*z2.... +ak*zk ]m1+m2+..+mk δ(z1 + z2 + .... –1 ) which is our final and most general result. Now if we set all mi = 1, we get 1/(a1a2 ....ak) = (k-1)! ∫dz1∫dz2....∫dzk 1/[ a1*z1+a2*z2.... +ak*zk ]k δ(z1 + z2 + .... –1 ) which is 8.59 on page 170 of BD.