Advanced Calculus (Solutions Manual) by Gerald B. Folland

By Gerald B. Folland

This can be the whole strategies handbook to Gerald Folland's Advanced Calculus.

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Improper Integrals 31 € ✚ ✄ ✎ € ☞❊➁❬✒ ➁ ✡✥✙▲❛ ✟ € ✎☞ ✍✑✏✓✒ ✎ ✟ ☞❊➁✫✒ ➁ ✍ ❵ ✌ ✍ ❵ ✌ , so On the other hand, ❽ ✝ ✝ € € € ✟✚ ✄❈❛ ✂ ☞✎✍✑✏✓✒ ✎ ✟ ✙▲❛ ➁ ✍ ▼❛ ✂ ✎ ✟ ☞❊➁❬✒ ➁❧⑩ ✍ ✙▲❛ ✟ ✂ ✎☞ ✍❀✙✜✒ ✎✢✜ ☞ ✙▲❛❜✒ ☞❊➁✫✒ ➁ ❽ ✍ ✌ ❵ ✌ ✍ ❵ ✌ ✍ € € ✎✢✜ ✎ ✟ ✄ ★▲❛ ✜ ✂ ☞❊➁✫✒ ➁⑧✍⑩❅▲❛ ✂ ☞❊➁✫✒ ➁✯✄ ❵ ✌ ❵ ✌ ✍ ✍ € ✎ ✟ ☞❊➁❬✒ ➁ ✟☎ ✡ ✚ ✟ ✎✢✜ ★▲❛ ✎ ✟ Thus, in ❽ ❽ , the terms involving ✍ add up to ✝ ✍ ❵ ✌ , and the terms involving ✍ add up to the negative of this quantity. ✂ ✁ ✐ € ✁ ☞❊✁✬✒ 8. 53a. Let ✁⑥✡✣✢ enough ❯ ➂✫✢■✒❁✍ ❵ ☞❊✁✿➂❬✢●✒✗ ❺✽✫ ☎ ❯ ✣✄ ❵ are all in a ball contained in .

B) ✌ ✝ ✌ ✪ € ✄✼ ✔ ✝ ✕ ✟ ❛ ✝✟ ✌ ✆ ✄✳✍❝❃ ✌ ✟ . ➁❦✄ ✰ ➁❋✄♠✰ (c) On the segment and ✌ ; on the segment ➇⑥✄✳ ✏ € from ➇⑥✄✖(0,0) ✰ € to (1,0) we have ➁ ✄❈➇ from➁✉✄(1,0)➇ to (1,1) and , and on the segment from (1,1) to (0,0) we have and ✌ we have ✌ ✌ . Hence ✰ ✌ ➇ ✡ ✝ ☞✎✍❀✙✜✒ ✌ ➁ ✡ ✝ € ☞❊➇ ✟ ✍ ✙▲➇❴✒ ✌ ➇❦✄✳✍❀✙❝✡ ✟✜ ✄✳✍ ✪✜ . the integral is ✝ ✄ ✁✳✄ ☛●☞❊❛❜✒ ✱✕q ❛★q ✗ ✁ ✌ ✂ ✡ ✄ ✡ ✝ ✞ ✁✲☞ ☛●☞❊❛❜✒❨✒✗ ☛ ☞❊❛❨✒✗ ✌ ❛ ✡ q by , .

6. Improper Integrals 31 € ✚ ✄ ✎ € ☞❊➁❬✒ ➁ ✡✥✙▲❛ ✟ € ✎☞ ✍✑✏✓✒ ✎ ✟ ☞❊➁✫✒ ➁ ✍ ❵ ✌ ✍ ❵ ✌ , so On the other hand, ❽ ✝ ✝ € € € ✟✚ ✄❈❛ ✂ ☞✎✍✑✏✓✒ ✎ ✟ ✙▲❛ ➁ ✍ ▼❛ ✂ ✎ ✟ ☞❊➁❬✒ ➁❧⑩ ✍ ✙▲❛ ✟ ✂ ✎☞ ✍❀✙✜✒ ✎✢✜ ☞ ✙▲❛❜✒ ☞❊➁✫✒ ➁ ❽ ✍ ✌ ❵ ✌ ✍ ❵ ✌ ✍ € € ✎✢✜ ✎ ✟ ✄ ★▲❛ ✜ ✂ ☞❊➁✫✒ ➁⑧✍⑩❅▲❛ ✂ ☞❊➁✫✒ ➁✯✄ ❵ ✌ ❵ ✌ ✍ ✍ € ✎ ✟ ☞❊➁❬✒ ➁ ✟☎ ✡ ✚ ✟ ✎✢✜ ★▲❛ ✎ ✟ Thus, in ❽ ❽ , the terms involving ✍ add up to ✝ ✍ ❵ ✌ , and the terms involving ✍ add up to the negative of this quantity. ✂ ✁ ✐ € ✁ ☞❊✁✬✒ 8. 53a. Let ✁⑥✡✣✢ enough ❯ ➂✫✢■✒❁✍ ❵ ☞❊✁✿➂❬✢●✒✗ ❺✽✫ ☎ ❯ ✣✄ ❵ are all in a ball contained in .

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