Green Function Homework: Problem & Solutions

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Homework Help Overview

The problem involves the application of the Leibniz rule in the context of differentiation under the integral sign, specifically related to a Green function problem. Participants are discussing the implications of treating certain functions as constants during differentiation.

Discussion Character

  • Conceptual clarification, Mathematical reasoning, Assumption checking

Approaches and Questions Raised

  • Participants explore whether an additional term is missing in the provided solutions and discuss the treatment of trigonometric functions as constants within the integral. Questions arise about the application of the Leibniz rule when reducing the integral to a single variable.

Discussion Status

The discussion is ongoing, with some participants providing references to external resources to clarify points. There is a mix of interpretations regarding the application of differentiation rules, and guidance has been offered about treating certain functions as constants.

Contextual Notes

Participants are working under the constraints of homework rules, which may limit the information they can share or the methods they can use. There is uncertainty about the completeness of the provided solutions and the assumptions made in the problem setup.

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Homework Statement



The problem is attached in the first picture, the provided solutions are in the second.

The Attempt at a Solution



I got to where they are, but aren't they missing an additional term of sin(t)*cos(∏)*f(∏) from the second integral in dx/dt ?
 

Attachments

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  • greenfunction2.jpg
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Millennial said:

Ok! I worked it out finally. Will the leibniz rule work if i take 'cos(t)' and 'sin(t)' out of the integral such that the inside of the integral only becomes one variable: ∫-sin(ζ)f(ζ) and ∫-cos(ζ)f(ζ) ??


The answer seems to use the leibniz rule but they imply that the 'cos(t)' and 'sin(t)' were taken out of the integral..
 
cos(t) and sin(t) are treated as constants in the integral because the integral is with respect to zeta, and the integral itself is treated as a constant in differentiation because the derivative is taken with respect to t. The function inside inside the integral that involves t is differentiated as usual, and the integral is treated as a constant so it is left as-is.
 
Millennial said:
cos(t) and sin(t) are treated as constants in the integral because the integral is with respect to zeta, and the integral itself is treated as a constant in differentiation because the derivative is taken with respect to t. The function inside inside the integral that involves t is differentiated as usual, and the integral is treated as a constant so it is left as-is.

Sorry I'm not sure what you mean but here's the method that I used (attached in the picture)

In the picture its ∫ f(x,t) from u to v.

In the question of this thread, I am above to take 'sin(t)' and 'cos(t)' out of the integration so that the integration only has 1 variable.

My question is:

How do we still apply (or can we) leibniz's rule if this is the case? (Only 1 variable)
 

Attachments

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Yes, you can, just take f(x,t)=f(t) (check the link I gave about differentiation under the integral sign.)
 

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