improper integrals notes - v: 2013-01-21 CALC 1501 LECTURE...

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v: 2013-01-21CALC 1501 LECTURE NOTESRASUL SHAFIKOV3.Improper Intergrals.So far we dealt with integration of continuous functions on bounded intervals. In this sectionwe will discuss integration of continuous functions on unbounded intervals, and also integration ofcertain unbounded functions.3.1.Unbounded intervals.Suppose a functionf(x) is continuous on the interval (a,), so thatthe integralRbaf(x)dxis well-defined for anyb > a. The limit of this integral asb→ ∞will becalled theimproper integral off(x)on(a,). That is(3.1)Zaf(x)dx= limb→∞Zbaf(x)dx.If the limit exists (i.e., it is a finite number), then we say that the integralRaf(x)dxconvergesoris convergent. If the limit is infinite, or does not exist, we say that the improper integraldivergesoris divergent.Example 3.1.Considerf(x) =11+x2. Then for anyb >0,Zb0dx1 +x2= tan-1xb0= tan-1b.Therefore,Z0dx1 +x2= limb→∞Zb0dx1 +x2= limb→∞tan-1b=π2.Thus, the integralR0dx1+x2converges toπ/2.Example 3.2.Considerf(x) =1xp, wherep >0 is a real number. Let us find the values of theexponentpfor which the integral(3.2)Z1dxxpconverges. Ifp6= 1, andb >1 any number, thenZb1dxxp=11-px1-pb1=11-p(b1-p-1).Ifp >1, thenb1-p0 asa→ ∞, and the integral in (3.2) converges. Ifp <1, thenb1-p→ ∞and the integral diverges. Suppose now thatp= 1. ThenZb1dxx= lnxb1= lnb,and sincelimb→∞lnb=, we conclude thatR1dxxdiverges.Thus, the improper integral in (3.2)converges forp >1 and diverges for 0< p1.1
2RASUL SHAFIKOVFigure 1.The graph ofe-xsinxIn a similar way we define improper integrals from-∞tob:Zb-∞f(x)dx=lima→-∞Zbaf(x)dx.Finally, we defineZ-∞f(x)dx=limb→ ∞a→-∞Zbaf(x)dx.In the latter case, we may also choose any numberAso thatZ-∞f(x)dx=ZA-∞f(x)dx+ZAf(x)dx.Then the integral on the left-hand side converges if and only if both integrals on the right-handside converge.Example 3.3.Determine whether the following integrals converge.(i)Z0-∞exsinx dx(ii)Z-∞dxx2-2x+ 2Solution: (i) Integrating by parts twice, we obtainZexsinx dx=12(-excosx+exsinx).Therefore,Z0-∞exsinx dx=lima-→∞12(-excosx+exsinx)0a=lima→-∞12(-1 +eacosa-easina) =-12.
CALC 1501 LECTURE NOTES3Indeed, since sinxand cosxare bounded functions, and limb→∞e-b= 0, it follows thatlimb→∞e-bsinb= limb→∞e-bcosb= 0.(ii) First note that denominator of the integrand does not vanish for anyx, and so the functionunder the integral sign is continuous onR. We haveZ-∞dxx2-2x+ 2=Z1-∞dxx2-2x+ 2+Z1dxx2-2x+ 2.

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