Physics 112An IntegralPeter Young(Dated: February 16, 2012)In class we showed that the determination of the specific heat of a degenerate Fermi gas involvedthe following integralI =Z∞−∞x2ex(ex+ 1)2dx .(1)The integrand can be written asxex/2+ e−x/22, (2)which is clearly an even function of x and so we can write I as an integral involving only positivevalues of x,I = 2Z∞0x2ex(ex+ 1)2dx . (3)This is not ex ac tl y of a standard type, but can be related to a more standard integral sinceI = −2dJ(a)daa=1,(4)whereJ(a) =Z∞0xeax+ 1dx=1a2Z∞0tet+ 1dt , (5)where, in the last line, we made the substitution ax = t. Hence, from Eqs. (4) and ( 5), we haveI = 4Z∞0tet+ 1dt ,(6)which is of a fairly standard type.We determine it, initially as a series, as follows:I = 4Z∞0tet+ 1dt= 4Z∞0te−t1 + e−tdt= 4Z∞0te−t− e−2t+ e−3t− e−4t+ · · · dt= 4Z∞0te−tdt1 −122+132−142+ · · ·, (7)2where, t o get the last line, we made the substitution t → t/2 in the second term, t → t/3 in thethird term, and so on . The integralR∞0te−tdt i s equal to 1! (= 1), and soI = 41 +132+152+ · · ·−122+142+162+ · · ·. (8)In the first set of rectangular brackets we include the missing terms (which i nvolve even integers)and then subtract them back out in the second term, i.e.I = 41 +122+132+142+ · · ·− 2122+142+162+ · · ·= 41 +122+132+142+ · · ·−121 +122+132+ · · ·= 41 −12 1 +122+132+142+ · · ·= 2 ζ(2) , (9)whereζ(2) ≡ 1 +122+132+142+ · · ·(10)is a zeta function and has value π2/6. (This will have been shown in 116C.) Hence, from Eqs. (9)and (10), we haveI =π23,(11)as st ate d in