Improper Integrls October 4, 7 Introduction We hve seen how to clculte definite integrl when the it is rel number. However, there re times when we re interested to compute the integrl sy for emple 3. Here we cn plot the grph nd see tht the region under the curve etends in the horizontl direction. Thus our notion of definite integrl tht we hve lerned before, does not pply in this cse. We cnnot just integrte it s usul nd plug in infinity. Integrls of these type re clled improper integrls. The second type of improper integrl is when the it intervl includes point where the integrnd is discontinous. There is specific wy to del with these kinds of integrls in generl. We will replce the infinity(or discontinuous point) with vrible (usully T ), do the integrl nd then tke the it of the result s T goes to infinity(discontinous point). Before tht, it is useful if we lern bout the bsic fundmentl of its of function, which include one-sided it nd lso L Hopitl s Rule since when we re deling with such integrls, there re times when we come cross of indeterminte form. Limits If function f() becomes rbitrrily close to single number L s pproches c from either side, then f() = L, c which is red s the it of f() s pproches c is L. Limits of function cn be esily evluted grphiclly. However most of the times, evluting the polynomil
functions re more fesible by using the direct substitution. Hence if f is polynomil function, nd c is rel number, then f() = f(c), c tht is simply done by plugging c into the function f(). Emple.. 3 (3 + ) 3. 4 5 3 4. 3 L Hopitl s Rule In order to clculte certin cses of its, we will mke use of L Hopitl s Rule when we reched n indeterminte form of prticulr cses. Generlly if then f() f() c g() = c g() = c f() c g() = f () c g () OR =, f () f () provided tht eists or =. It is importnt to note tht we re c g () c g () not doing the quotient rule differentition here. But we re differentiting denomintor nd numertor seprtely. Emple Find the it of the following epressions using L Hopitl s Rule. + 6 + 9. 3 + 3 8. 4 3. 3 5 6 + 8 4. π sin cos
5. π cos π 6. e ln 7. 8. + ln( ) ( ) Sometimes, the first derivtive still give us n indeterminte form. We cn just pply L Hopitl s rule gin. Emple 3 Find the it of the following epressions using L Hopitl s Rule. 4. 5 6 3. p e p + e p cos p 3. e ln t 4. t cosec t 5. tn sin 6. b ln b b 7. cos + 3 4 4 One-sided Limit Lter in improper integrl, we will use one sided it to see the vlue of the function prroches from the left or right. This type of behvior cn be described more concisely with the concept of one-sided it. f() = L c c + f() = L Limit from the left Limit from the right Emple 4 3
3. 3. + 5 3. + + 5 4. + 5. + 5 Improper Integrls of Type I : Infinite Integrls. Suppose tht f() is defined nd continous for ll,. We define T f() = f(). T. In the sme wy, if f() is defined nd continous in n intervl, we define f() = T T f(). 3. If f() is defined nd continous for ll then f() = c f() + c f(), where c is ny number(we usully use zero for simplicity of evluting the integrl). For Cse nd Cse, the improper integrls re sid to converge if the it eist. If the it does not eist, the improper integrl is sid to diverge. In Cse 3, the improper integrl is sid to converge if both improper integrls on the right hnd side converge. Otherwise, it is sid to be diverge if either one or both of the improper integrls on the right diverge. Emple 5 Determine if ech of the following integrls converge or diverge. If it converges, determine its vlue. 4
.. 3. 4. 5. 6. e ( ) ( + ) + e + e 7. 8. 9. (tn - ) + ( + ), given substitution u = +. e e, given substitution u = e. 6 Improper Integrls of Type II : Infinite Discontinuity This type of integrls is wht we lwys encounter when we re doing definite integrls. It is just so hppened, before, most of the definite integrls tht we worked on in previous chpter, do not hve discontinuity. However, from now on we hve to set our mind tht everytime we re given definite integrl to evlute, we should check if it is n improper integrl of infinite discontinuity.. If f() is continous nd defined in the intervl < b nd hs infinite discontinuity t = b, then T f() = f(). T b 5
. In the sme wy, if f() is continous nd defined in the intervl < b nd hs discontinuity t =, then f() = T + T f(). 3. If f() is continous in the intervl b, ecept for point t = c in the intervl (, b) where f() hs discontinuity, then f() = c f() + c f(). For Cse nd, the improper integrl is sid to converge if its it eists. the it does not eist, the improper integrl is sid to diverge. For Cse 3, the integrl on the left hnd side is sid to converge if both improper integrls on the right converge. If either one or both diverge,then the improper integrl is diverge. Emple 6 Evlute ech of the following integrls if the integrls converge... 3. 4. 5. 6. 7. 8. 9.. 3 3 3 4 e π 3 ( ) 3 ( 3) ln sin cos 6 If
7 Conclusion This chpter is just continution of previous chpter. Thus from now on, you should hve set your mind tht if you were given definite integrl, try to check if it is Type or Type, then solve ccordingly. Tht mens, if the integrtion hs discontinuities or if the it of the integrtion is infinity, then you should know how to rewrite the integrl in product form. Additionlly, once you solve the integrtion nd you insert the it but ended up with n indeterminte form of or, then you should know tht you cn use L Hopitl s Rule. 7