Binomial integrals- plus solutions than Chevished All binomial are A: I=∫xMdx(xN+k)P We are going to take profit from the series shmx·chnx, and also from the circulars sinusmx·cosnx, and the analog shmx/chnx chnx/shmx, and the circulars sinusmx/cosnx and cosnx/sinusmx. This integrands are solved at Wikipedia (when m and n are integers, then also our problem.) WE will need that N would be 2, then we shall do x=t2/N dx=t(2-N)/Ndt=dt·t(2/N-1)
I=∫dt·t(2/N-1)t2M/N(t2+k2)P Now the blue
braket can have the forms (+/-t2+/-k2)p the changes will be t=kshw (when two +) or t=kchw (if +and-) or t=k·sinus(w) (if -&+) and the result I=∫dw·ch(w)sh[2(M+1)/N-1(w)[ch(w)]2P=∫dwshmwchnw m=[2(M+1)/N1] n=2P+1 than have to be integers- (case first) for the second I=∫dwch[2(M+1)/N-1](w)[sh(w)]2P+1 and the third I=∫dw·sinus[2(M+1)/N-1](w)[(cos(w)]2P+1 Then what we need is that m and n would be integers or what is the same P-1 and m+1 A) first case m=2(M+1)/N n=2P P can be multiple of ½ (3/2, 5/2 7/2…) besides an integer, equal than (M+1)/N; then more cases than Chebishev. B) m=2P n=2(M+1)/N as before C) n=2P m=2(M+1)/N no new exigences. The three cases have the solutions what said Chebishev, plus all the multiples of 1/2. Besides, one of the exponents m or n, can be <0. When one exponent is negative we have other formulas:
A)
B): similarly
sinm/cosn, and cosn/sinm that can be seen solved in wikipedia
For products of hyperbolics they are integrated in wikipedia. ( see at the end of the anex): I(m,n)= ∫du·shmu·chnu= =shm-1u·chn+1u/(m+n)-(m-1)/(n+1) ∫shm-2u·chnu·du 2P+1 or n=2(M+1)/N-1 as we had seen
(Chebyshev believed to have demonstrated that there was no more cases, than his three integers [P, (M+1)/N, P+(M+1)/N] but they can be also multiples of 1/2.