This document provides a detailed analysis of improper integrals and vector functions, exploring their properties and behavior. From the study of real numbers to the application of vector analysis, this comprehensive analysis offers insights into the world of mathematical functions.
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[...] Finally: sur or on a . Thus: for all On From, according to the triangle inequality: Or on a : . We therefore obtain: Moreover, we have also previously demonstrated that for all From inequality: So we can also deduce: On - First case: if , the first inequality of the previous question gives us then: - If , The second inequality of the previous question then gives us: Finally, for all Let it be the following integral: with belonging to . [...]
[...] And as, according to the previous question, the improper integral is convergent, we deduce that the integral is convergent as the sum of two convergent integrals. Sur , on Let it be the function defined on by: . On The derivative is always negative so is decreasing. As , we deduce that is negative about . Finally: sur Be it the function defined on by: . On The derivative is always negative so is decreasing. As , we deduce that is negative about . [...]
[...] So is integrable on a neighborhood of to the right if and only if what is equivalent to - Study in to the left: We have, when tend towards : So is integrable on a neighborhood of to the left if and only if In conclusion, is integrable on if and only if II. Exercise 3 When tend vers , on a . Thus, when tend vers : From where: being a continuous function on , prolongable by continuity in , on deduce that the improper integral is convergent. Improper integral is convergent. Or on for all de : On deduces that the improper integral is absolutely convergent and therefore convergent. [...]
