Indeterminate Forms, Improper Integrals, Taylor, and Maclaurin Series PDF

Gandhinagar Institute of Technology Lecture Notes Of INDETERMINATE FORMS IMPROPER INTEGRALS. TAYLOR’S SERIES AND MACLAURIN’S SERIES Program (UG): B. TECH...

Gandhinagar Institute of Technology Lecture Notes Of INDETERMINATE FORMS IMPROPER INTEGRALS. TAYLOR’S SERIES AND MACLAURIN’S SERIES Program (UG): B. TECH Name of Department: Mathematics and Humanities Academic Year: 2023-2024 Semester: 1st Subject Code: 10000101 Subject Name: Engineering Mathematics I Name of the Institute: Gandhinagar Institute of Technology Mathematics and Humanities Department 10000101 EM I  Gandhinagar Institute of Technology INDETERMINANT FORMS Introduction We have studied certain rules to evaluate the limits. But some limits cannot be evaluated by using these rules. These limits are known as indeterminate forms.These are seven types of indeterminate forms. 0 ∞ (1) (2) (3) 0 × ∞ (4) ∞ − ∞ 0 ∞ (5) 1∞ (6) ∞0 (7) 00 These limits can be evaluated by using L’Hospital’s rule. L’Hospital’s Rule If 𝑓(𝑥) and 𝑔(𝑥) are two functions of 𝑥 which can be expanded by Taylor’s series in thE neighbourhood of 𝑥 = 𝑎 and if lim 𝑓(𝑥) = 𝑓(𝑎) = 0, lim 𝑔(𝑥) = 𝑔(𝑎) = 0, 𝑥→𝑎 𝑥→𝑎 then 𝑓(𝑥) 𝑓 ′ (𝑥) lim = lim ′ 𝑥→𝑎 𝑔(𝑥) 𝑥→𝑎 𝑔 (𝑥) Note The following standard limits can be used to solved the problems: sin 𝑥 tan 𝑥 𝑎𝑥 −1 (1) lim =1 (2) lim =1 (3) lim = log 𝑎 𝑥→0 𝑥 𝑥→0 𝑥 𝑥→0 𝑥 𝑒 𝑥 −1 1 1 𝑥 (4) lim =1 (5) lim (1 + 𝑥)𝑥 = 𝑒 (6) lim (1 + ) = 𝑒 𝑥→0 𝑥 𝑥→0 𝑥→∞ 𝑥 sin−1 𝑥 sinh 𝑥 tan−1 𝑥 (7) lim =1 (8) lim =1 (9) lim =1 𝑥→0 𝑥 𝑥→0 𝑥 𝑥→0 𝑥 𝟎 Type 1 : Form 𝟎 Problems under this type are solved by using L’Hospital’s rule considering the fact that 𝑓(𝑥) 𝑓′ (𝑥) lim = lim if lim 𝑓(𝑥) = 0 and lim 𝑔(𝑥) = 0 𝑥→𝑎 𝑔(𝑥) 𝑥→𝑎 𝑔′ (𝑥) 𝑥→𝑎 𝑥→𝑎 Example – 1 𝒙 𝒆𝒙 −𝐥𝐨𝐠(𝟏+𝒙) Evaluate 𝐥𝐢𝐦. 𝒙→𝟎 𝒙𝟐 Solution: 𝑥 𝑒 𝑥 −log(1+𝑥) 0 Let 𝐼 = lim [ form] 𝑥→0 𝑥2 0 1 𝑒 𝑥+ 𝑥 𝑒 𝑥− 0 = lim 1+𝑥 [ form] [Applying L’Hospital’s rule] 𝑥→0 2𝑥 0 1 𝑒 + 𝑒 𝑥 + 𝑥 𝑒 𝑥 + (1+𝑥)2 𝑥 = lim [Applying L’Hospital’s rule] 𝑥→0 2 3 = 2 Mathematics and Humanities Department 10000101 EM I  Gandhinagar Institute of Technology Example – 2 𝒆𝒙 + 𝒆−𝒙 − 𝒙𝟐 − 𝟐 Evaluate 𝐥𝐢𝐦. 𝒙→𝟎 𝐬𝐢𝐧𝟐 𝒙 −𝒙𝟐 Solution: 𝑒 𝑥 + 𝑒 −𝑥 − 𝑥 2 − 2 0 Let 𝐼 = lim [ form] 𝑥→0 sin2 𝑥 − 𝑥 2 0 𝑒 𝑥 − 𝑒 −𝑥 − 2𝑥 0 = lim [ form] [Applying L’Hospital’s rule] 𝑥→0 2 sin x cos x − 2𝑥 0 𝑒 𝑥 − 𝑒 −𝑥 − 2𝑥 = lim 𝑥→0 sin 2𝑥 − 2𝑥 𝑒 𝑥 + 𝑒 −𝑥 − 2 0 = lim [ form] [Applying L’Hospital’s rule] 𝑥→0 2 cos 2𝑥 − 2 0 𝑒 𝑥 − 𝑒 −𝑥 0 = lim [ form] [Applying L’Hospital’s rule] 𝑥→0 −4 sin 2𝑥 0 𝑒 𝑥 + 𝑒 −𝑥 0 = lim [ form] [Applying L’Hospital’s rule] 𝑥→0 −8 cos 2𝑥 0 1+1 1 = = − −8 4 Example – 3 𝒙 − 𝒙𝒙 Evaluate 𝐥𝐢𝐦. 𝒙→𝟏 𝟏 + 𝐥𝐨𝐠 𝒙 − 𝒙 Solution: 𝑥 − 𝑥𝑥 0 Let 𝐼 = lim [ form] 𝑥→1 1 + log 𝑥 − 𝑥 0 𝑥 − 𝑒 𝑥 log 𝑥 = lim 𝑥→1 1 + log 𝑥 − 𝑥 1 − 𝑒 𝑥 log 𝑥 (1 + log 𝑥) 0 = lim 1 [ form] [Applying L’Hospital’s rule] 𝑥→1 −1 0 𝑥 𝑥 log 𝑥 1 −𝑒 (1 + log 𝑥)2 − 𝑒 𝑥 log 𝑥 ( ) 0 = lim 1 𝑥 [ form][Applying L’Hospital’s 𝑥→1 − 2 0 𝑥 rule] = 2 Example – 4 𝐜𝐨𝐬 𝟐 𝝅𝒙 Evaluate 𝐥𝐢𝐦𝟏. 𝒆𝟐𝒙 − 𝟐𝒙𝒆 𝒙→ 𝟐 Solution: cos2 𝜋𝑥 0 Let 𝐼 = lim1 [ form] 𝑥→ 𝑒 2𝑥 − 2𝑥𝑒 0 2 2 cos 𝜋𝑥 (−𝜋 sin 𝜋𝑥) = lim1 [Applying L’Hospital’s rule] 𝑥→ 2𝑒 2𝑥 − 2𝑒 2 −𝜋 sin 2𝜋𝑥 0 = lim1 [ form] 𝑥→ 2(𝑒 2𝑥 − 𝑒) 0 2 Mathematics and Humanities Department 10000101 EM I  Gandhinagar Institute of Technology −2𝜋2 cos 2𝜋𝑥 = lim1 [Applying L’Hospital’s rule] 𝑥→ 4 𝑒 2𝑥 2 𝜋2 = 2𝑒 Exercises: 𝑥 2 +2 cos 𝑥 − 2 (1) Evaluate lim. [Ans: 0] 𝑥→ 0 𝑥 sin 𝑥 𝑒 𝑥 sin 𝑥 − 𝑥 − 𝑥 2 2 (2) Evaluate lim. [Ans: − ] 𝑥→ 0 𝑥 2 + 𝑥 log(1−𝑥) 3 ∞ Type 2 : Form ∞ Problems under this type are solved by using L’Hospital’s rule considering the fact that 𝑓(𝑥) 𝑓′ (𝑥) lim = lim if lim 𝑓(𝑥) = ∞ and lim 𝑔(𝑥) = ∞ 𝑥→𝑎 𝑔(𝑥) 𝑥→𝑎 𝑔′ (𝑥) 𝑥→𝑎 𝑥→𝑎 Example – 1 𝝅 𝐥𝐨𝐠(𝒙− ) 𝟐 Prove that 𝐥𝐢𝐦𝝅 = 𝟎. 𝒙→ 𝐭𝐚𝐧 𝒙 𝟐 Solution: 𝜋 log(𝑥− ) ∞ 2 Let 𝐼 = lim𝜋 [ form] 𝑥→ tan 𝑥 ∞ 2 1 𝜋) ( 𝑥− = lim 2 𝜋 sec2 𝑥 [Applying L’Hospital’s rule] 𝑥→ 2 cos2 𝑥 0 = lim𝜋 𝜋 [ form] 𝑥→ 𝑥 − 0 2 2 2 cos 𝑥 (− sin 𝑥) = lim𝜋 [Applying L’Hospital’s rule] 𝑥→ 1 2 = 0 Mathematics and Humanities Department 10000101 EM I  Gandhinagar Institute of Technology Example – 2 Evaluate 𝐥𝐢𝐦 𝐥𝐨𝐠 𝐭𝐚𝐧 𝒙 𝐭𝐚𝐧 𝟐𝒙. 𝒙→𝟎 Solution: Let 𝐼 = lim log tan 𝑥 tan 2𝑥 𝑥→0 log tan 2𝑥 ∞ = lim [ form] 𝑥 → 0 log tan 𝑥 ∞ 1 ∙ 2 sec2 2𝑥 = lim tan 2𝑥 1 [Applying L’Hospital’s rule] 𝑥→0 ∙sec2 𝑥 tan 𝑥 tan 𝑥 ∙ sec2 2𝑥 𝑥 = lim tan 2𝑥 𝑥 → 0 2𝑥 ∙ sec2 𝑥 tan 𝑥 = 1 [∵ lim = 1] 𝑥→0 𝑥 Exercises (1) Prove that lim log 𝑥 tan 𝑥 = 1. 𝑥→0 log (𝑥 − 𝑎) (2) Prove that lim = 1. 𝑥 → 𝑎 log (𝑎𝑥 − 𝑎𝑎 ) log(1−𝑥) (3) Prove that lim = 0. 𝑥→1 cot 𝜋𝑥 Type 3 : 𝟎 × ∞ Form To solve the problems of the type lim [𝑓(𝑥) ∙ 𝑔(𝑥)], when lim 𝑓(𝑥) = 0 and lim 𝑔(𝑥) = ∞ (i.e. 0 × ∞ 𝑥→𝑎 𝑥→𝑎 𝑥→𝑎 form) 𝑓(𝑥) 𝑔(𝑥) We write lim [𝑓(𝑥) ∙ 𝑔(𝑥)] = lim 1 or lim 1 𝑥→𝑎 𝑥→𝑎 𝑔(𝑥) 𝑥→𝑎 𝑓(𝑥) 0 ∞ These new forms are of the type or respectively, which can be solved 0 ∞ using L’Hospital’s rule. Example – 1 Prove that 𝐥𝐢𝐦 𝐬𝐢𝐧 𝒙 𝐥𝐨𝐠 𝒙 = 𝟎. 𝒙→𝟎 Solution: Let 𝐼 = lim sin 𝑥 log 𝑥 [0 × ∞ form] 𝑥→0 log 𝑥 ∞ = lim [ form] 𝑥→0 cosec 𝑥 ∞ 1 = lim 𝑥 [Applying L’Hospital’s rule] 𝑥→0 −cosec 𝑥 cot 𝑥 tan 𝑥 = − lim sin 𝑥 ∙ 𝑥→0 𝑥 Mathematics and Humanities Department 10000101 EM I  Gandhinagar Institute of Technology tan 𝑥 = − lim sin 𝑥 ∙ lim 𝑥→0 𝑥→0 𝑥 tan 𝑥 = 0 [∵ lim = 1] 𝑥→0 𝑥 Example – 2 1 Prove that lim (𝑎 𝑥 − 1) 𝑥 = log 𝑎. 𝑥→∞ Solution: 1 Let 𝑙 = lim (𝑎 𝑥 − 1) 𝑥 [0 × ∞ form] 𝑥→∞ 1 (𝑎 𝑥 −1) 0 = lim 1 [ form] 𝑥→∞ 0 𝑥 1 Putting = 𝑡, when 𝑥 → ∞, 𝑡 → 0 𝑥 𝑎𝑡 −1 0 𝑙 = lim [ form] 𝑡→0 𝑡 0 𝑎𝑡 log 𝑎 = lim [Applying L’Hospital’s rule] 𝑡→0 1 = log 𝑎 Exercises 𝑥 1 (1) Evaluate lim log (2 − ) cot(𝑥 − 𝑎). [Ans : − ] 𝑥→𝑎 𝑎 𝑎 (2) Prove that tan 𝑥 log 𝑥 = 0. Type 4 : ∞ − ∞ Form To evaluate the limits of the type lim [𝑓(𝑥) − 𝑔(𝑥)], when lim 𝑓(𝑥) = ∞ and 𝑥→𝑎 𝑥→𝑎 0 lim 𝑔(𝑥) = ∞ [ i.e. (∞ − ∞) form], we reduce the expression in the form of 𝑥→𝑎 0 ∞ or by taking LCM or by rearranging the terms and then apply L’Hospital’s ∞ rule. Example – 1 𝟏 𝟏 Evaluate 𝐥𝐢𝐦 [ − 𝟐 𝐥𝐨𝐠(𝟏 + 𝒙)]. 𝒙→𝟎 𝒙 𝒙 Solution: 1 1 Let 𝑙 = lim [ − log(1 + 𝑥)] [∞ − ∞ form] 𝑥→0 𝑥 𝑥2 𝑥 − log (1+𝑥) 0 = lim [ ] [ form] 𝑥→0 𝑥2 0 Mathematics and Humanities Department 10000101 EM I  Gandhinagar Institute of Technology 1 1− 0 = lim [ 1+𝑥 ] [ form] [Applying L’Hospital’s 𝑥→0 2𝑥 0 rule] 1 2 ( ) = lim [ 1+𝑥 ] [Applying L’Hospital’s rule] 𝑥→0 2 1 = 2 Example – 2 𝒙 𝟏 Evaluate 𝐥𝐢𝐦 ( − ). 𝒙→𝟏 𝒙−𝟏 𝐥𝐨𝐠 𝒙 Solution: 𝑥 1 Let 𝑙 = lim ( − ) [∞ − ∞ form] 𝑥→1 𝑥−1 log 𝑥 𝑥 log 𝑥 − 𝑥 + 1 0 = lim [ ] [ form] 𝑥→1 log 𝑥 (𝑥−1) 0 1+ log 𝑥 − 1 = lim [ 𝑥−1 ] [Applying L’Hospital’s rule] 𝑥→1 + log 𝑥 𝑥 log 𝑥 0 = lim [ 𝑥−1 ] [ form] 𝑥→1 + log 𝑥 0 𝑥 1 = lim [ 1 𝑥 1 ] [Applying L’Hospital’s rule] 𝑥→1 + 𝑥2 𝑥 1 = 2 Example – 3 𝟏 𝟏 Evaluate 𝐥𝐢𝐦 ( 𝟐 − ). 𝒙→𝟎 𝒙 𝐬𝐢𝐧𝟐 𝒙 Solution: 1 1 Let 𝐼 = lim ( 2 − 2 ) [∞ − ∞ form] 𝑥→0𝑥 sin 𝑥 sin2 𝑥 − 𝑥 2 0 = lim ( ) [ form] 𝑥→0 𝑥 2 sin2 𝑥 0 sin2 𝑥 − 𝑥 2 sin2 𝑥 = lim ( ) [∵ lim = 1] 𝑥→0 𝑥4 𝑥→0 𝑥2 2 sin 𝑥 cos 𝑥− 2𝑥 = lim ( ) [Applying L’Hospital’s rule] 𝑥→0 4𝑥 3 sin 2𝑥− 2𝑥 = lim ( ) 𝑥→0 4𝑥 3 2 cos 2𝑥− 2 = lim ( ) [Applying L’Hospital’s rule] 𝑥→0 12𝑥 2 −4 sin 2𝑥 = lim ( ) [Applying L’Hospital’s rule] 𝑥→0 24𝑥 −8 cos 2𝑥 = lim ( ) [Applying L’Hospital’s rule] 𝑥→0 24 8 = − 24 Mathematics and Humanities Department 10000101 EM I  Gandhinagar Institute of Technology 1 = − 3 Exercises 1 1 1 (1) Evaluate lim [ − ]. [Ans: − ] 𝑥→2 𝑥−2 log(𝑥−1) 2 𝜋 𝜋 𝜋2 (2) Prove that lim [ − ]=. 𝑥→0 4𝑥 2𝑥 (𝑒 𝜋𝑥 + 1) 8 Type – 5 : 𝟏∞ , ∞𝟎 , 𝟎𝟎 Forms To evaluate the limits of the type lim [𝑓(𝑥)]𝑔(𝑥) , which takes any one of the 𝑥→𝑎 ∞ 0 0 form 1 , ∞ , 0 for 𝑓(𝑥) > 0, we proceed as follows: Let 𝑙 = lim [𝑓(𝑥)]𝑔(𝑥) where 𝑓(𝑥) > 0 𝑥→𝑎 log 𝑙 = lim [𝑔(𝑥) ∙ log 𝑓(𝑥)] 𝑥→𝑎 Which takes the form ∞ × 0, i.e., type 3 form. Example – 1 𝟏 Prove that 𝐥𝐢𝐦 (𝐚𝐱 + 𝐱)𝐱 = 𝐚𝐞. 𝐱→𝟎 Solution: 1 Let 𝑙 = lim (𝑎 𝑥 + 𝑥)𝑥 [1∞ form] 𝑥→0 1 log 𝑙 = lim ∙ log(𝑎 𝑥 + 𝑥) [∞ × 0 form] 𝑥→𝑎 𝑥 log(𝑎𝑥 +𝑥) 0 log 𝑙 = lim [ form] 𝑥→𝑎 𝑥 0 1 𝑥 (𝑎𝑥 log 𝑎+1) log 𝑙 = lim 𝑎 +𝑥 [Applying L’Hospital’s rule] 𝑥→𝑎 1 𝑎0 log 𝑎+1 log 𝑙 = 𝑎0 + 0 log𝑒 𝑎+log𝑒 𝑒 = 1 = log 𝑎𝑒 Hence, 𝑙 = 𝑎𝑒 Example – 2 𝟏 𝟏𝒙 + 𝟐 𝒙 + 𝟑𝒙 𝒙 Evaluate 𝐥𝐢𝐦 ( ). 𝒙→𝟎 𝟑 Solution: 1 1𝑥 + 2𝑥 + 3𝑥 𝑥 Let 𝑙 = lim ( ) [1∞ form] 𝑥→0 3 1 1𝑥 + 2𝑥 + 3𝑥 log 𝑙 = lim log ( ) 𝑥→0 𝑥 3 1𝑥 + 2𝑥 + 3𝑥 log( ) 0 3 = lim [ form] 𝑥→0 𝑥 0 Mathematics and Humanities Department 10000101 EM I  Gandhinagar Institute of Technology 3 1𝑥 log 1 + 2𝑥 log 2 + 3𝑥 log 3 = lim ∙ [Applying L’Hospital’s 𝑥→0 1𝑥 + 2𝑥 + 3𝑥 3 rule] 1 = (log 2 + log 3) 3 1 = log 6 3 1 = log(6)3 1 log 𝑙 = (6)3 Example – 3 Evaluate 𝐥𝐢𝐦 (𝐜𝐨𝐬 𝒙)𝐜𝐨𝐭 𝒙. 𝒙→𝟎 Solution: Let 𝑙 = lim (cos 𝑥)cot 𝑥 [1∞ form] 𝑥→0 log 𝑙 = lim cot 𝑥 ∙ log(cos 𝑥) 𝑥→0 log(cos 𝑥) 0 = lim [ form] 𝑥→0 tan 𝑥 0 1 (− sin 𝑥) log 𝑙 = lim cos 𝑥 [Applying L’Hospital’s rule] 𝑥→0 sec2 𝑥 = 0 Hence, 𝑙 = 𝑒 0 = 1 Example – 4 𝟏 𝐭𝐚𝐧 𝒙 Evaluate 𝐥𝐢𝐦 ( ). 𝒙 𝒙→𝟎 Solution: 1 tan 𝑥 Let 𝑙 = lim ( ) [∞0 form] 𝑥 𝑥→0 1 log 𝑙 = lim tan 𝑥 log ( ) 𝑥→0 𝑥 log 𝑥 ∞ = lim − [ form] 𝑥→0 cot 𝑥 ∞ 1 (− ) = lim 𝑥 [Applying L’Hospital’s rule] 𝑥→0 − cosec2 𝑥 1 𝑥 = lim 1 𝑥→0 sin2 𝑥 sin2 𝑥 = lim 𝑥→0 𝑥 sin2 𝑥 = lim ∙𝑥 𝑥→0 𝑥2 sin 𝑥 2 = lim ( ) ∙ lim 𝑥 𝑥→0 𝑥 𝑥→0 =1∙0 Mathematics and Humanities Department 10000101 EM I  Gandhinagar Institute of Technology = 0 Hence, 𝑙 = 𝑒 0 = 1 Example – 5 𝟏 Prove that 𝐥𝐢𝐦 (𝟏 − 𝒙𝟐 )𝐥𝐨𝐠(𝟏−𝒙) = 𝒆. 𝒙→𝟏 Solution: 1 2 )log(1−𝑥) Let 𝑙 = lim (1 − 𝑥 [00 form] 𝑥→1 1 ∞ log 𝑙 = lim log(1 − 𝑥 2 ) [ form] 𝑥→1 log(1−𝑥) ∞ 2𝑥 − 2 (1 − 𝑥 ) log 𝑙 = lim 1 [Applying L’Hospital’s rule] 𝑥→1 (−1) (1−𝑥) 2𝑥 (1−𝑥) = lim 𝑥→1 (1−𝑥) (1+𝑥) 2𝑥 = lim 𝑥→1 1+𝑥 = 1 Hence, 𝑙 = 𝑒 Example – 6 𝝅 −𝒙 Evaluate 𝐥𝐢𝐦𝝅 (𝐜𝐨𝐬 𝒙) 𝟐. 𝒙→ 𝟐 Solution: 𝜋 Let 𝑙 = lim𝜋 (cos 𝑥)2 − 𝑥 [00 form] 𝑥→ 2 𝜋 log 𝑙 = lim𝜋 ( − 𝑥) log cos 𝑥 [0 × ∞ form] 2 𝑥→ 2 log cos 𝑥 ∞ = lim𝜋 1 [ form] 𝑥→ 𝜋 ∞ 2 ( − 𝑥) 2 sin 𝑥 − cos 𝑥 = lim 𝜋 1 𝑥→ − 𝜋 2 (−1) 2 ( − 𝑥) 2 − tan 𝑥 = lim𝜋 1 𝑥→ 𝜋 2 2 ( − 𝑥) 2 𝜋 2 = lim𝜋 − ( − 𝑥) tan 𝑥 𝑥→ 2 2 𝜋 2 ( − 𝑥) 2 = lim𝜋 − 𝑥→ cot 𝑥 2 Mathematics and Humanities Department 10000101 EM I  Gandhinagar Institute of Technology 𝜋 −2 ( − 𝑥) (−1) 2 = lim𝜋 𝑥→ − cosec2𝑥 2 𝜋 = lim𝜋 2 ( – 𝑥) sin2 𝑥 𝑥→ 2 2 log 𝑙 = 0 Hence, 𝑙 = 𝑒 0 = 1 Exercises 1 − 𝑥𝑥 (1) Evaluate lim. 𝑥→0 𝑥 log 𝑥 [Ans: −1] 1 1−cos 𝑥 (2) Prove that lim ( ) = 1. 𝑥 𝑥→0 (3) Evaluate lim (cot 𝑥)sin 𝑥 𝑥→0 [Ans: 1] 2 (4) Evaluate lim𝜋 (cosec 𝑥)tan 𝑥. 𝑥→ 2 1 [Ans: 𝑒 ] 2 1 1 𝑎𝑥 + 𝑏𝑥 + 𝑐 𝑥 3𝑥 (5) Prove that lim ( ) = (𝑎𝑏𝑐)9 𝑥→0 3 Mathematics and Humanities Department 10000101 EM I  Gandhinagar Institute of Technology IMPROPER INTEGRALS Introduction 𝑏 The definition of a definite integral ∫𝑎 𝑓(𝑥)𝑑𝑥 requires the interval [𝑎, 𝑏] be finite. The fundamental theorem of calculus requires that 𝑓(𝑥) be continuous on [𝑎, 𝑏] or at least bounded. In this chapter, we will study a method of evaluating integrals that fail these requirements either because their limits of integration are infinite, or because a finite number of dis continuities exist on the interval [𝑎, 𝑏]. Integrals that fail either of these requirements are known as improper integrals. Improper integrals cannot be computed using a normal Riemann integral. Improper Integrals 𝑏 The integral ∫𝑎 𝑓(𝑥)𝑑𝑥 is called an improper integral if (1) one or both limits of integration are infinite (2) function 𝑓(𝑥) becomes infinite at a point within or at the end points of the interval of integration. Improper integrals are classified into three kinds. Improper Integrals of the First Kind It is a definite integral in which one or both limits of integration are infinite. For ∞ example, ∫0 𝑒 −𝑥 𝑑𝑥 is an improper integral of the first kind since the upper limit of integration is infinite. These integrals are evaluated as follows: (1) If 𝑓(𝑥) is continuous on [𝑎, ∞) then ∞ 𝑏 ∫𝑎 𝑓(𝑥) 𝑑𝑥 = lim ∫𝑎 𝑓(𝑥) 𝑑𝑥 ……(1) 𝑏→∞ (2) If 𝑓(𝑥) is continuous on (−∞, 𝑏] then 𝑏 𝑏 ∫−∞ 𝑓(𝑥) 𝑑𝑥 = lim ∫𝑎 𝑓(𝑥) 𝑑𝑥 ……(2) 𝑎→−∞ (3) If 𝑓(𝑥) is continuous on (−∞, ∞] then ∞ 𝑏 ∫−∞ 𝑓(𝑥) 𝑑𝑥 = 𝑎→−∞ lim ∫𝑎 𝑓(𝑥) 𝑑𝑥 𝑏→∞ 0 𝑏 = lim ∫𝑎 𝑓(𝑥) 𝑑𝑥 + lim ∫0 𝑓(𝑥) 𝑑𝑥 ……(3) 𝑎→−∞ 𝑏→∞ The improper integral is said to converge (or exist) when the limit in RHS of (1), (2) and (3) exist (or finite). Otherwise, it is said to diverge. Example – 1 ∞ 𝟏 Evaluate ∫𝟏 𝒅𝒙. √𝒙 Solution: Mathematics and Humanities Department 10000101 EM I  Gandhinagar Institute of Technology ∞ 𝑏 1 1 ∫ 𝑑𝑥 = lim ∫ 𝑑𝑥 √𝑥 𝑏→∞ √𝑥 1 1 = lim (2√𝑥)1𝑏 𝑏→∞ = lim (2√𝑏 − 2) 𝑏→∞ =∞ Example – 2 ∞ 𝒅𝒙 Evaluate ∫𝟎 𝟐. 𝒙 +𝟏 Solution: ∞ 𝑏 𝑑𝑥 𝑑𝑥 ∫ 2 = lim ∫ 2 𝑥 + 1 𝑏→∞ 𝑥 + 1 0 0 = lim (tan−1 𝑥)𝑏0 𝑏→∞ = lim (tan−1 𝑏 − tan−1 0) 𝑏→∞ = lim tan−1 𝑏 𝑏→∞ = tan−1 ∞ 𝜋 = 2 Example – 3 ∞ 𝒅𝒗 Evaluate ∫𝟎 (𝟏+𝒗𝟐). (𝟏+𝐭𝐚𝐧−𝟏 𝒗) Solution: 1 ∞ 𝑑𝑣 𝑏 (1+𝑣2 ) ∫0 (1+𝑣 2) (1+tan−1 𝑣) = lim ∫0 𝑑𝑣 𝑏→∞ (1+tan−1 𝑣) = lim |log|1 + tan−1 𝑣||𝑏0 𝑏→∞ = lim [log |1 + tan−1 𝑏| − log 1] 𝑏→∞ = lim 𝑙𝑜𝑔(1 + 𝑡𝑎𝑛−1 ∞) − 0 𝑏→∞ 𝜋 = log (1 + ) 2 Example – 4 ∞ 𝟏 Evaluate ∫−∞ 𝒅𝒙. 𝟏+𝒙𝟐 Solution: ∞ 1 0 1 𝑏 1 ∫−∞ 1+𝑥 2 𝑑𝑥 = lim ∫𝑎 1+𝑥 2 𝑑𝑥 + lim ∫0 1+𝑥 2 𝑑𝑥 𝑎→−∞ 𝑏→∞ = lim | tan−1 𝑥|𝑎0 + lim | tan−1 𝑥|0𝑏 𝑎→−∞ 𝑏→∞ = lim (0 − tan−1 𝑎) + lim (tan−1 𝑏 − 0) 𝑎→−∞ 𝑏→∞ 𝜋 𝜋 = + 2 2 Mathematics and Humanities Department 10000101 EM I  Gandhinagar Institute of Technology =𝜋 Example – 5 𝟎 Evaluate ∫−∞ 𝒆𝟐𝒙 𝒅𝒙. Solution: 0 0 ∫−∞ 𝑒 2𝑥 𝑑𝑥 = lim ∫𝑎 𝑒 2𝑥 𝑑𝑥 𝑎→−∞ 0 𝑒 2𝑥 = lim | | 𝑎→−∞ 2 𝑎 1 1 = lim ( − 𝑒 2𝑎 ) 𝑎→−∞ 2 2 1 = −0 2 1 = 2 Improper Integrals of the Second Kind It is a definite integral in which integrand become infinite (or unbounded or discontinuous) At one or more points within or at the end points of the interval of integration, for example, 11 1 (1) ∫0 𝑑𝑥 is an improper integral of the second kind as is not continuous at 𝑥 = 𝑥 𝑥 0. 2 1 1 (2) ∫−2 2 𝑑𝑥 is an improper integral of the second kind because 2 is not 𝑥 −1 𝑥 −1 continuous at 𝑥 = −1 and 𝑥 = 1. These integrals are evaluated as follows: (1) If 𝑓(𝑥) is unbounded at 𝑥 = 𝑎 then 𝑏 𝑏 ∫𝑎 𝑓(𝑥) 𝑑𝑥 = lim ∫𝑐 𝑓(𝑥) 𝑑𝑥 …….(1) 𝑐→𝑎 (2) If 𝑓(𝑥) is unbounded at 𝑥 = 𝑏 then 𝑏 𝑐 ∫𝑎 𝑓(𝑥) 𝑑𝑥 = lim ∫𝑎 𝑓(𝑥) 𝑑𝑥 …….(2) 𝑐→𝑏 (3) If 𝑓(𝑥) is unbounded at 𝑥 = 𝑎 and 𝑥 = 𝑏 then 𝑏 0 𝑐2 ∫𝑎 𝑓(𝑥) 𝑑𝑥 = lim ∫𝑐 𝑓(𝑥) 𝑑𝑥 + lim ∫0 𝑓(𝑥) 𝑑𝑥 ………(3) 𝑐1 →𝑎 1 𝑐2 →𝑏 The improper integral is said to converge (or exist) when the limit in RHS of (1), (2) and (3) exist (or finite). Otherwise, it is said to diverge. Example – 1 𝟑 𝟏 Evaluate ∫𝟎 𝒅𝒙. √𝟑−𝒙 Solution: 1 The integrand is unbounded at 𝑥 = 3. √3−𝑥 3 1 𝑐 1 ∫0 √3−𝑥 𝑑𝑥 = lim ∫0 𝑑𝑥 𝑐→3 √3−𝑥 Mathematics and Humanities Department 10000101 EM I  Gandhinagar Institute of Technology = lim (−2 √3 − 𝑥)𝑐0 𝑐→3 = lim (−2 √3 − 𝑐 + 2√3) 𝑐→3 = 2 √3 Example – 2 𝟑 𝒅𝒙 Check the convergence of ∫𝟎. √𝟗 −𝒙𝟐 Solution: 1 The integrand is unbounded at 𝑥 = 3. √9 − 𝑥 2 3 𝑑𝑥 𝑐 𝑑𝑥 ∫0 = lim ∫0 √9 −𝑥 2 𝑐→3 √9 −𝑥 2 𝑐 −1 𝑥 = lim (sin ) 𝑐→3 3 0 𝑐 = lim (sin−1 − sin−1 0) 𝑐→3 3 −1 −1 = sin 1 − sin 0 𝜋 = −0 2 𝜋 = (finite) 2 Hence, the improper integral is cgt. Example – 3 𝟑 𝒅𝒙 Evaluate ∫𝟎 𝟐. (𝒙 − 𝟏)𝟑 Solution: 1 The integrand 2 is unbounded at 𝑥 = 1. (𝑥 − 1)3 3 𝑑𝑥 𝑐 𝑑𝑥 3 𝑑𝑥 ∫0 2 = lim ∫0 1 2 + lim ∫𝑐 2 (𝑥 − 1)3 𝑐1 →1 (𝑥 − 1)3 𝑐2 →1 2 (𝑥 − 1)3 1 𝑐1 1 3 (𝑥 − 1)3 (𝑥 − 1)3 = lim ( 1 ) + lim ( 1 ) 𝑐1 →1 3 𝑐2 →1 3 0 𝑐2 1 1 1 1 = lim 3 [(𝑐1 − 1) − (−1) ] + lim 3 [(3 − 1)3 − (𝑐2 − 1)3 ] 3 3 𝑐1 →1 𝑐2 →1 1 1 = 3 [0 − (−1) ] + 3 [2 − 0] 3 3 1 = 3 [23 + 1] Example – 4 𝟓 𝟏 Evaluate ∫𝟎 𝟐 𝒅𝒙. 𝒙 Solution: 1 The integrand 2 is unbounded at 𝑥 = 0. 𝑥 Mathematics and Humanities Department 10000101 EM I  Gandhinagar Institute of Technology 5 5 1 1 ∫ 2 𝑑𝑥 = lim ∫ 2 𝑑𝑥 𝑥 𝑐→0 𝑥 0 𝑐 1 5 = lim (− ) 𝑐→0 𝑥 𝑐 1 1 = lim [− + ] 𝑐→0 5 𝑐 1 =− +∞ 5 =∞ Example – 5 𝟏 𝒅𝒙 Check the convergence of ∫𝟎. 𝟏−𝒙 Solution: 1 The integrand is unbounded at 𝑥 = 1. 1−𝑥 1 𝑐 𝑑𝑥 𝑑𝑥 ∫ = lim ∫ 1−𝑥 𝑐→1 1−𝑥 0 0 = lim | − log(1 − 𝑥)|𝑐0 𝑐→1 = lim [− log(1 − 𝑐) + log 1] 𝑐→1 = − log(1 − 1) =∞ Hence, the integrand is divergent. Improper Integrals of the Third Kind It is a definite integral in which one or both limits of integration are infinite, and the integrand become infinite at one or more points within or at the end points of the interval of integration. Thus, it is a combination of the first kind and the second kind. ∞ 1 For example, ∫0 2 𝑑𝑥 is an improper integral of the third kind as the upper limit 𝑥 1 of integration is infinite and integrand 2 is infinite at 𝑥 = 0. 𝑥 Example – 1 ∞ 𝟏 Evaluate the improper integral ∫𝟎 𝟐 𝒅𝒙. 𝒙 Solution: ∞ 1 𝑐2 1 1 1 ∫ 𝑑𝑥 = lim ∫ 𝑑𝑥 + lim ∫ 𝑑𝑥 𝑥2 𝑐1 →0 𝑥2 𝑐2 →∞ 𝑥2 0 𝑐1 1 1 1 1 𝑐2 = lim |− | + lim |− | 𝑐1 →0 𝑥 𝑐1 𝑐2 →∞ 𝑥 1 Mathematics and Humanities Department 10000101 EM I  Gandhinagar Institute of Technology 1 1 = lim (−1 + ) + lim (− + 1) 𝑐1 →0 𝑐 1 𝑐 𝑐2 →∞ 2 =∞+1 =∞ 𝑏 1 = lim ∫1 2 𝑑𝑥 𝑏→∞𝑥 1 𝑏 = lim |− | 𝑏→∞ 𝑥 1 1 = lim (− + 1) 𝑏→∞ 𝑏 =1 ∞ 1 Thus, ∫1 2 𝑑𝑥 is convergent. 𝑥. Mathematics and Humanities Department 10000101 EM I  Gandhinagar Institute of Technology Taylor’s series and Maclaurin’s Series Definition: If 𝑓(𝑥) is an infinitely differentiale function of 𝑥 throught some interval containing ′𝑎′ as an interior point, then 𝑓(𝑥) can be expanded as power series in (𝑥 − 𝑎). This series is called Taylor’s series generated by 𝑓 at 𝑥 = 𝑎. (𝑥 − 𝑎) (𝑥 − 𝑎)2 (𝑥 − 𝑎)3 𝑓(𝑥) = 𝑓(𝑎) + 𝑓′(𝑎) + 𝑓′′(𝑎) + 𝑓′′′(𝑎) + ⋯ 1! 2! 3! Putting 𝑥 − 𝑎 = ℎ we get, ℎ ℎ2 ℎ3 𝑓(𝑎 + ℎ) = 𝑓(𝑎) + 𝑓′(𝑎) + 𝑓′′(𝑎) + 𝑓′′′(𝑎) + ⋯ 1! 2! 3! Putting 𝑎 = 0 we get, 𝑥 𝑥2 𝑥3 𝑓(𝑥) = 𝑓(0) + 𝑓′(0) + 𝑓′′(0) + 𝑓′′′(0) + ⋯ 1! 2! 3! This series is known as Maclaurin’s series. This is nothing but the Taylor’s series at origin. We can find the expansion of different functions as follows: 𝑥 𝑥2 𝑥3 𝑥4 𝑓(𝑥) = 𝑒 = 1 + 𝑥 + + + + ⋯ 2! 3! 4! 𝑥3 𝑥5 𝑓(𝑥) = sin 𝑥 = 𝑥 − + + ⋯ 3! 5! 𝑥2 𝑥4 𝑓(𝑥) = cos 𝑥 = 1 − + + ⋯ 2! 4! 𝑥2 𝑥3 𝑥4 𝑓(𝑥) = log(1 + 𝑥) = 𝑥 − + − + ⋯ 2 3 4 Example-1: Expand 𝐥𝐨𝐠 𝒙, in powers of (𝒙 − 𝟏) and hence evaluate 𝐥𝐨𝐠 𝟏. 𝟏. Solution: We have 𝑓(𝑥) = log 𝑥 and 𝑎 = 1 𝑓(𝑥) = log 𝑥 𝑓(1) = 0 1 𝑓′(1) = 1 𝑓′(𝑥) = 𝑥 1 𝑓′′(1) = −1 𝑓′′(𝑥) = − 2 𝑥 2 𝑓′′′(1) = 2 𝑓′′′(𝑥) = 3 𝑥 Mathematics and Humanities Department 10000101 EM I  Gandhinagar Institute of Technology 6 𝑓′′′′(1) = −6 𝑓 ′′′′(𝑥) = − 𝑥4 Substituting these values in Taylor’s series, (𝑥 − 𝑎) (𝑥 − 𝑎)2 (𝑥 − 𝑎)3 𝑓(𝑥) = 𝑓(𝑎) + 𝑓′(𝑎) + 𝑓′′(𝑎) + 𝑓′′′(𝑎) + ⋯ 1! 2! 3! (𝑥 − 1)2 (𝑥 − 1)3 (𝑥 − 1)4 log 𝑥 = 0 + (𝑥 − 1)(1) + (−1) + (2) + (−6) … 2! 3! 4! (𝑥 − 1)2 (𝑥 − 1)3 (𝑥 − 1)4 log 𝑥 = (𝑥 − 1) − + − … 2 3 4 Now putting 𝑥 = 1.1, we get 1 1 1 log 1.1 = 0.1 − (0.1)2 + (0.1)3 − (0.1)4 + ⋯ 2 3 4 log 1.1 = 0.1 − 0.005 + 0.000333 − 0.000025 + ⋯ log 1.1 = 0.095308 Example:2 Expand 𝟑𝒙𝟑 + 𝟖𝒙𝟐 + 𝒙 − 𝟐, in powers of (𝒙 − 𝟑). Solution: We have 𝑓(𝑥) = 3𝑥 3 + 8𝑥 2 + 𝑥 − 2 and 𝑎 = 3 𝑓(𝑥) = 3𝑥 3 + 8𝑥 2 + 𝑥 − 2 𝑓(3) = 154 ′(𝑥) 2 𝑓′(3) = 130 𝑓 = 9𝑥 + 16𝑥 + 1 𝑓 ′′(𝑥) = 18𝑥 + 16 𝑓′′(3) = 70 𝑓′′′(𝑥) = 18 𝑓′′′(3) = 18 Substituting these values in Taylor’s series, (𝑥 − 𝑎) (𝑥 − 𝑎)2 (𝑥 − 𝑎)3 𝑓(𝑥) = 𝑓(𝑎) + 𝑓′(𝑎) + 𝑓′′(𝑎) + 𝑓′′′(𝑎) + ⋯ 1! 2! 3! 1 1 𝑓(𝑥) = 154 + (𝑥 − 3)(130) + (𝑥 − 3)2 (70) + (𝑥 − 3)3 (18) 2! 3! 𝑓(𝑥) = 154 + 130(𝑥 − 3) + 35(𝑥 − 3)2 + 3(𝑥 − 3)3 𝝅 Example: 3 Expand 𝐬𝐢𝐧 ( + 𝒙) in powers of 𝒙. Hence find the value of 𝟒 𝐬𝐢𝐧 𝟒𝟒 and 𝐬𝐢𝐧 𝟒𝟔. Solution: 𝜋 𝑓(𝑥 + 𝑎) = sin ( + 𝑥) 4 𝜋 ∴ 𝑓(𝑥) = sin 𝑥 & 𝑎 = 4 Mathematics and Humanities Department 10000101 EM I  Gandhinagar Institute of Technology 𝜋 𝜋 ∴ sin ( + 𝑥) = 𝑓 (𝑥 + ) 4 4 𝜋 𝜋 ′ 𝜋 𝑥 2 ′′(𝜋) 𝑥 3 ′′′(𝜋) 𝑥 4 ′′′(𝜋) ∴ sin ( + 𝑥) = 𝑓 ( ) + 𝑥𝑓 ( ) + 𝑓 4 + 𝑓 4 + 𝑓 4 + ⋯ 4 4 4 2! 3! 4! 𝑓(𝑥) = sin 𝑥 𝜋 1 𝑓( ) = 4 √2 𝑓 ′(𝑥) = cos 𝑥 𝜋 1 𝑓′ ( ) = 4 √2 𝑓 ′′(𝑥) = − sin 𝑥 𝜋 ′′( ) 1 𝑓 4 =− √2 𝑓 ′′′(𝑥) = − cos 𝑥 𝜋 ′′′( ) 1 𝑓 4 =− √2 Now from the above result, 𝜋 1 1 𝑥2 1 𝑥3 1 sin ( + 𝑥) = +𝑥 + (− ) + (− ) + ⋯ 4 √2 √2 2! √2 3! √2 2 3 𝜋 1 𝑥 𝑥 ∴ sin ( + 𝑥) = (1 + 𝑥 − − + ⋯ ) 4 √2 2! 3! Now putting 𝜋 3.14 𝑥 = −1 = − = − = −0.0174 180 180 In above result, 1 (0.0174)2 (0.0174)3 sin 44 = [1 − 0.0174 − + + ⋯ ] = 0.6947 √2 2! 3! Now putting 𝜋 3.14 𝑥=1= = = 0.0174 180 180 In above result, 1 (0.0174)2 sin 46 = [1 − 0.0174 − … ] = 0.7193 √2 2! Mathematics and Humanities Department 10000101 EM I  Gandhinagar Institute of Technology Example: 3 Using Taylor’s theorem, evaluate up to four places of decimals √𝟏. 𝟎𝟐. Solution: Let 𝑓(𝑥) = √𝑥 𝑓(𝑥 + ℎ) = √𝑥 + ℎ By Taylor's series, ′ ℎ2 ′′ ℎ3 ′′′ 𝑓(𝑥 + ℎ) = 𝑓(𝑥) + ℎ𝑓 (𝑥) + 𝑓 (𝑥) + 𝑓 (𝑥) + ⋯ 2! 3! (i) Putting 𝑥 = 1, ℎ = 0.02, 𝑓(𝑥 + ℎ) = √𝑥 + ℎ = √1 + 0.02 = √1.02 ′ (0.02)2 ′′ = 𝑓(1) + (0.02)𝑓 (1) + 𝑓 (1) + ⋯ 2! 𝑓(𝑥) = √𝑥, 𝑓(1) = 1 1 1 𝑓 ′ (𝑥) = , 𝑓 ′ (1) = 2 √𝑥 2 1 1 𝑓 ′′ (𝑥) = − 3 , 𝑓 ′′ (1) = − 4 4𝑥 2 and so on 1 (0.02)2 1 √1.02 = 1 + (0.02) ( ) + (− ) 2 2! 4 = 1.0099 approx. Mathematics and Humanities Department 10000101 EM I