1 = 1 two F1 1 k, k; 1 2k; five k four 2 1 (1 ) two (3 k ) (1 k)two F1 2 k,1 k; 1 2k
1 = 1 2 F1 1 k, k; 1 2k; five k four two 1 (1 ) 2 (3 k ) (1 k)2 F1 2 k,1 k; 1 2k; 2 1 1 k; 1 2k; 2 1 . (189) (1 k)(two k)2 F1 three k,The hypergeometric functions can be expressed with regards to elementary functions utilizing Equation (A16), leading to-dz F k, 3 k; 1 2k; – 2 1 k two 1 (1 z ) 1 z=2k2 k (6 five) 4 . 3/2 2k 1 (1 ) 2k (2 k )(1 1 )(190)Substituting the above result into Equation (185), we arrive atE V P 0 , j j (-1) j1 cosh two 0 cosh 2 0 = j 6 sinh4 0 j =1d(1 – 2 two )1 12k42k2 k (6 5) P. (1 )3/2 1 (191)E In order to extract the higher temperature limit of V P , to begin with we ought to account for ,its divergence when 1. This can be achieved by altering the integration variable in Equation (191) from to j 0 = sinh2 , (192) 2 exactly where was introduced in Equation (186). We obtainE V P ,j j (-1) j1 cosh 2 0 cosh two 0 = 12 (1 – 2 ) sinh4 j0 j =11d sinhj 02kj sinh2 2j4j 6 sinh2 2 0 five k sinh (sinh2 j0 )3/22k2 sinh2 2 0 j 0 P. j two sinh2 2 0 (193)Due to the fact we’re interested only GYY4137 Purity & Documentation inside the small 0 behaviour, it is tempting to carry out a series expansion with the integrand with respect to 0 then integrate order by order. We require j terms up to 4 to balance the sinh4 two 0 issue within the denominator with the factor BI-0115 site appearingSymmetry 2021, 13,43 ofin front of the integral. On the other hand, the greater order powers of 0 also introduce damaging powers of , producing this process invalid. Even though the full integral cannot be computed analytically, we initial note that P (183) admits the following series expansion with respect to 0 : eight j 0 4 2 two P = 3- (194) (1 – )(1 – ) O( six ). 0 9 two The above expansion shows that the dependence of P on comes in only at the fourth order with respect to 0 . It is actually not too difficult to observe that the second and third terms j in Equation (193) contribute terms of order sinh two 0 , therefore for these terms, the O( 4 ) con0 tribution to P is usually neglected. We therefore carry out the computation in two methods. Initially, we approximate P 3 (corresponding to its 0 = 0 limit). The integral might be performed by j switching the integration variable to = / sinh2 2 0 and noting that sinh2 j 0(sinhj 0 -2 2 )d1 12k42k2 k(six 5) 3/2 1 (1 )= 4e- jM0 1 k j 0 tanh . 2(195)The contribution on the fourth order term in P may be estimated by setting 0 = 0 within the integrand, giving1d sinhj 02k4j sinh2 2 0 j 6 sinh2 two 0 5 k sinh (sinh2 j0 )3/2 two 32 2 j 0 42k2 sinh2 two 0 j 0 (P – three) j 2 sinh2 two 0 j-d(1 – )(1 – )=-Substituting the above results into Equation (193) yieldsE V P 0 ,16j 0(three – ).(196)j =(-1) j1 coshj 0 j 0 two cosh 2 (1 – 2 ) sinh4 j0e- jM0 1 k j 0 tanh 2-4j 0(three – ) ,(197)where the O( six ) contributions were ignored inside the square brackets. We now carry out a 0 tiny 0 expansion inside the summand and then calculate the sum term by term to findE V P = ,1-4 2 7 four T0 2 T0 R 3 – 9MT0 (3) 6M2 2 45 18- MT0 ln two R 4M2 32 6- O( T0 1 ) . (198)1 177 two 17 -294 540M2 two R – 45M2 R – R2 6480 6Symmetry 2021, 13,44 ofUsing the volume integral of your SC provided in Equation (129), it could be noticed that the volume E E SC integral V0 , = three V P M V , from the power density might be written as four four ,0E V0 , = V RKT ,E1-2 two T0 R two 24-MT0 2 ln 2- O( T0 1 ) . (199)1 17 177 two R – 90M2 R – R2 -294 360M2 2 8640 6ECompared to the RKT prediction V RKT derived in Equation (53), you will discover cor,two rections appearing already at order O( T0 ). The higher temperature limit obtained in Equation (199) is validated in Figure ten by comparison with numerically obtained re3 E E SC sults for V0 , =.
