Some Useful Integrals of Exponential Functions


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1 prvious indx nxt Som Usful Intgrls of Exponntil Functions Michl Fowlr W v shown tht diffrntiting th xponntil function just multiplis it by th constnt in th xponnt, tht is to sy, d x x Intgrting th xponntil function, of cours, hs th opposit ffct: it divids by th constnt in th xponnt: x 1 s you cn sily chck by diffrntiting both sids of th qution An importnt dfinit intgrl (on with limits) is x, 0 x 1 Notic th minus sign in th xponnt: w nd n intgrnd tht dcrss s x gos towrds infinity, othrwis th intgrl will itslf b infinit To visuliz this rsult, w plot blow x nd 3x Not tht th grn lin forms th hypotnus of rightngld tringl of r 1, nd it is vry plusibl from th grph tht th totl r undr th x curv is th sm, tht is, 1, s it must b Th 3x curv hs r 1/3 undr it, ( 3)
2 Now for somthing bit mor chllnging: how do w vlut th intgrl x I? ( hs to b positiv, of cours) Th intgrl will dfinitly not b infinit: it flls off qully fst in both positiv nd ngtiv dirctions, nd in th positiv dirction for x grtr thn 1, it s smllr thn x, which w know convrgs To s bttr wht this function looks lik, w plot it blow for 1 (rd) nd 4 (blu)
3 3 Notic first how much fstr thn th ordinry xponntil x this function flls wy Thn not tht th blu curv, 4, hs bout hlf th totl r of th 1 curv In fct, th r gos s 1/ Th grn lins hlp s tht th r undr th rd curv (positiv plus ngtiv) is somwht lss thn, in fct it s 177 pproximtly But it s not so sy to vlut! Thr is trick: squr it Tht is to sy, writ ( ) x y I dy Now, this product of two intgrls long lins, th xintgrl nd th yintgrl, is xctly th sm s n intgrl ovr pln, th (x, y) pln, strtching to infinity in ll dirctions W cn rwrit it x y x y r dy dy dy whr r x + y, r is just th distnc from th origin (0, 0) in th (x, y) pln Th pln is dividd up into tiny squrs of r dy, nd doing th intgrl mounts to finding th vlu of r for ch tiny squr, multiplying by th r of tht squr, nd dding th contributions from vry squr in th pln In fct, though, this pproch is no sir thn th originl problm th trick is to notic tht r th intgrnd hs circulr symmtry: for ny circl cntrd t th origin (0, 0), it hs th sm vlu nywhr on th circl To xploit this, w shouldn t b dividing th (x, y) pln up
4 4 into littl squrs t ll, w should b dividing it into rgions hving ll points th sm distnc from th origin Ths r clld nnulr rgions: th r btwn two circls, both cntrd t th origin, th innr on of rdius r, th slightly biggr outr on hving rdius r + dr W tk dr to b vry smll, so this is thin circulr strip, of lngth r (th circumfrnc of th circl) nd brdth dr, nd thrfor its totl r is rdr (nglcting trms lik dr, which bcom ngligibl for dr smll nough) So, th contribution from on of ths nnulr rgions is ovr th whol pln is: r rdr, nd th complt intgrl ( I ) 0 r rdr This intgrl is sy to vlut: mk th chng of vribl to u r, du rdr giving so tking squr roots u I du ( ) 0 x I
5 5 Som Intgrls Usful in th Kintic Thory of Gss W cn sily gnrt mor rsults by diffrntiting I() bov with rspct to th constnt! Diffrntiting onc: so w hv x x d 1 d x d d x 1 x nd diffrntiting this rsult with rspct to givs 4 x x 3 4 Th rtio of ths two intgrls coms up in th kintic thory of gss in finding th vrg kintic nrgy of molcul with Mxwll s vlocity distribution 4 x x x x Finding this rtio without doing th intgrls: It is intrsting to not tht this rtio could hv bn found with much lss work, in fct without vluting th intgrls fully, s follows: Mk th chng of vribl y x, so dy nd 3/ x y 3/ x y dy C whr C is constnt indpndnt of, bcus hs compltly dispprd in th intgrl ovr y (Of cours, w know C /, but tht took lot of work) Now, th intgrl with x 4 in plc of x is givn by diffrntiting th x intgrl with rspct to, nd multiplying by 1, s discussd bov, so, diffrntiting th right hnd sid of th bov qution, th x 4 intgrl is just ( 3/) C 5/, nd th C cncls out in th rtio of th intgrls
6 6 Howvr, w do nd to do th intgrls t on point in th kintic thory: th ovrll normliztion of th vlocity distribution function is givn by rquiring tht 1 f( v) dv 4 A v mv /kt nd this in fct dtrmins A, using th rsults w found bov, giving 3/ m mv /kt f() v 4 v kt On lst trick W didn t nd this in th kintic thory lctur, but is sms pity to rviw xponntil intgrls without mntioning it It s sy to do th intgrl It cn b writtn x bx I b (, ) x b b /4 b /4 x b b /4 ( / ) ( / ) I, b / whr to do th lst stp just chng vribls from x to y x  b/ This cn vn b usd to vlut for xmpl x cosbx by writing th cosin s sum of xponntils prvious indx nxt
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