9 De nite integrals using the residue theorem

1 Topic 9 NotesJeremy Orloff9 Definite integrals using the residue IntroductionIn this topic we ll use the residue theorem to compute some real definite integrals . baf(x)dxThe general approach is always the same1. Find a complex analytic functiong(z) which either equalsfon the real axis or whichis closely connected tof, (x) = cos(x),g(z) = Pick a closed contourCthat includes the part of the real axis in the The contour will be made up of pieces. It should be such that we can compute g(z)dzover each of the pieces except the part on the real Use the residue theorem to compute Cg(z) Combine the previous steps to deduce the value of the integral we integrals of functions that decayThe theorems in this section will guide us in choosing the closed contourCdescribed in first theorem is for functions that decay faster than 1 (a)Supposef(z) is defined in the upper half-plane.

2 If there is ana >1andM >0 such that|f(z)|<M|z|afor|z|large thenlimR CRf(z)dz= 0,whereCRis the semicircle shown below on the (z)Im(z)R RCRRe(z)Im(z)R RCR19 DEFINITE integrals using THE residue THEOREM2 Semicircles: left:Rei , 0< < right:Rei , < <2 .(b)Iff(z) is defined in the lower half-plane and|f(z)|<M|z|a,wherea >1 thenlimR CRf(z)dz= 0,whereCRis the semicircle shown above on the prove (a), (b) is essentially the same. We use the triangle inequality for integralsand the estimate given in the hypothesis. ForRlarge CRf(z)dz CR|f(z)||dz| CRM|z|a|dz|= 0 MRaRd =M Ra >1 this clearly goes to 0 asR . QEDThe next theorem is for functions that decay like 1/z. It requires some more care to stateand (a)Supposef(z) is defined in the upper half-plane.

3 If there is anM >0such that|f(z)|<M|z|for|z|large then fora >0limx1 ,x2 C1+C2+C3f(z)eiazdz= 0,whereC1+C2+C3is the rectangular path shown below on the (z)Im(z)x1 x2i(x1+x2)C1C2C3Re(z)Im(z)x1 x2 i(x1+x2)C1C2C3 Rectangular paths of height and widthx1+x2.(b)Similarly, ifa <0 thenlimx1 ,x2 C1+C2+C3f(z)eiazdz= 0,whereC1+C2+C3is the rectangular path shown above on the contrast to theorem this theorem needs to include the factor (a)We start by parametrizingC1,C2, : 1(t) =x1+it,tfrom 0 tox1+x29 DEFINITE integrals using THE residue THEOREM3C2: 2(t) =t+i(x1+x2),tfromx1to x2C3: 3(t) = x2+it,tfromx1+x2to we look at each integral in turn. We assumex1andx2are large enough that|f(z)|<M|z|on each of the curvesCj.

4 C1f(z)eiazdz C1|f(z)eiaz||dz| C1M|z||eiaz||dz|= x1+x20M x21+t2|eiax1 at|dt Mx1 x1+x20e atdt=Mx1(1 e a(x1+x2)) >0, it is clear that this last expression goes to 0 asx1andx2go to . C2f(z)eiazdz C2|f(z)eiaz||dz| C2M|z||eiaz||dz|= x1 x2M t2+ (x1+x2)2|eiat a(x1+x2)|dt Me a(x1+x2)x1+x2 x1+x20dt Me a(x1+x2)Again, clearly this last expression goes to 0 asx1andx2go to .The argument forC3is essentially the same as forC1, so we leave it to the proof for part (b) is the same. You need to keep track of the sign in the exponentialsand make sure it is Example below for an example using theorem integrals and 0 Example 1(1 +x2) :Letf(z) = 1/(1 +z2) DEFINITE integrals using THE residue THEOREM4It is clear that forzlargef(z) 1 particular, the hypothesis of theorem is satisfied.

5 using the contour shown belowwe have, by the residue theorem , C1+CRf(z)dz= 2 i residues offinside the contour.(1)Re(z)Im(z)R RCRC1iWe examine each of the pieces in the above equation. CRf(z)dz: By theorem (a),limR CRf(z)dz= 0. C1f(z)dz: Directly, we see thatlimR C1f(z)dz= limR R Rf(x)dx= f(x)dx= lettingR , Equation 1 becomesI= f(x)dx= 2 i residues offinside the , we compute the needed residues :f(z) has poles of order 2 at i. Onlyz=iisinside the contour, so we compute the residue there. Letg(z) = (z i)2f(z) =1(z+i) (f,i) =g (i) = 2(2i)3=14iSo,I= 2 iRes(f,i) = 1x4+ DEFINITE integrals using THE residue THEOREM5 Solution:Letf(z) = 1/(1 +z4). We use the same contour as in the previous exampleRe(z)Im(z)R RCRC1ei /4ei3 /4As in the previous example,limR CRf(z)dz= 0andlimR C1f(z)dz= f(x)dx= , by the residue theoremI= limR C1+CRf(z)dz= 2 i residues offinside the poles offare all simple and atei /4,ei3 /4,ei5 /4,ei7 ei /4and ei3 /4are inside the contour.

6 We compute their residues as limits usingL Hospital s rule. Forz1= ei /4:Res(f,z1) = limz z1(z z1)f(z) = limz z1z z11 +z4= limz z114z3=14ei3 /4=e i3 /44and forz2= ei3 /4:Res(f,z2) = limz z2(z z2)f(z) = limz z2z z21 +z4= limz z214z3=14ei9 /4=e i /44So,I= 2 i(Res(f,z1) + Res(f,z2)) = 2 i( 1 i4 2+1 i4 2)= 2 i( 2i4 2)= 22 Example >0. Show 0cos(x)x2+b2dx= e :The first thing to note is that the integrand is even, soI=12 cos(x)x2+ DEFINITE integrals using THE residue THEOREM6 Also note that the square in the denominator tells us the integral is absolutely have to be careful because cos(z) goes to infinity in either half-plane, so the hypothesesof theorem are not satisfied. The trick is to replace cos(x) by eix, so I= eixx2+b2dx,withI=12Re( I).

7 Now letf(z) =eizz2+ +iywithy >0 we have|f(z)|=|ei(x+iy)||z2+b2|=e y|z2+b2|.Since e y<1,f(z) satisfies the hypotheses of theorem in the upper half-plane. Nowwe can use the same contour as in the previous examplesRe(z)Im(z)R RCRC1ibWe havelimR CRf(z)dz= 0andlimR C1f(z)dz= f(x)dx= , by the residue theorem I= limR C1+CRf(z)dz= 2 i residues offinside the poles offare at biand both are simple. Onlybiis inside the contour. We computethe residue as a limit using L Hospital s ruleRes(f,bi) = limz bi(z bi)eizz2+b2=e , I= 2 iRes(f,bi) = e ,I=12Re( I) = e b2b,as :Be careful when replacing cos(z) by eizthat it is appropriate. A key point inthe above example was thatI=12Re( I). This is needed to make the replacement DEFINITE integrals using THE residue Trigonometric integralsThe trick here is to put together some elementary properties ofz= ei on the unit e i = 1 cos( ) =ei + e i 2=z+ 1 sin( ) =ei e i 2i=z 1 start with an example.

8 After that we ll state a more general 2 0d 1 +a2 2acos( ).Assume that|a|6= :Notice that [0,2 ] is the interval used to parametrize the unit circle asz= ei .We need to make two substitutions:cos( ) =z+ 1/z2dz=iei d d =dzizMaking these substitutions we getI= 2 0d 1 +a2 2acos( )= |z|=111 +a2 2a(z+ 1/z)/2 dziz= |z|=11i((1 +a2)z a(z2+ 1)) , letf(z) =1i((1 +a2)z a(z2+ 1)).The residue theorem impliesI= 2 i residues offinside the unit can factor the denominator:f(z) = 1ia(z a)(z 1/a).The poles are ata,1/a. One is inside the unit circle and one is |a|>1 then 1/ais inside the unit circle and Res(f,1/a) =1i(a2 1)If|a|<1 thenais inside the unit circle andRes(f,a) =1i(1 a2)9 DEFINITE integrals using THE residue THEOREM8We haveI={2 a2 1if|a|>12 1 a2if|a|<1 The example illustrates a general technique which we state (x,y) is a rational function with no poles on the circlex2+y2= 1then forf(z) =1izR(z+ 1/z2,z 1/z2i)we have 2 0R(cos( ),sin( ))d = 2 i residues offinside|z|= make the same substitutions as in Example So, 2 0R(cos( ),sin( ))d = |z|=1R(z+ 1/z2,z 1/z2i)dzizThe assumption about poles means thatfhas no poles on the contour|z|= 1.}

9 The residuetheorem now implies the Integrands with branch cutsExample 0x1/31 + :Letf(x) =x1/31 + this is asymptotically comparable tox 5/3, the integral is absolutely convergent. Asa complex functionf(z) =z1/31 +z2needs a branch cut to be analytic (or even continuous), so we will need to take that intoaccount with our choice of , choose the following branch cut along the positive real axis. That is, forz=rei noton the axis, we have 0< <2 .Next, we use the contourC1+CR C2 Crshown DEFINITE integrals using THE residue THEOREM9Re(z)Im(z)CRC1 C2 Cri iContour around branch cut: inner circle of radiusr, outer of put convenient signs on the pieces so that the integrals are parametrized in a naturalway.

10 You should read this contour as havingrso small thatC1andC2are essentially onthex-axis. Note well, that, sinceC1andC2are on opposite sides of the branch cut, theintegral C1 C2f(z)dz6= we analyze the integral over each piece of the : theorem says thatlimR CRf(z)dz= : For concreteness, assumer <1/2. We have|z|=r, so|f(z)|=|z1/3||1 +z2| r1/31 r2 (1/2)1/33 the last number in the above equationM. We have shown that, for smallr,|f(z)|< , Crf(z)dz 2 0|f(rei )||irei |d 2 0 Mrd = 2 this goes to zero asr :limr 0,R C1f(z)dz= 0f(x)dx= : We have (essentially) = 2 , soz1/3= ei2 /3|z|1/3. Thus,limr 0,R C2f(z)dz= ei2 /3 0f(x)dx= ei2 poles off(z) are at i. Sincefis meromorphic inside our contour the residue theoremsays C1+CR C2 Crf(z)dz= 2 i(Res(f,i) + Res(f, i)).