Extremely difficult double integral.

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Homework Help Overview

The problem involves evaluating a double integral of the form \(\int_{0}^{8} \int_{y^{1/3}}^{2} \frac{1}{x^4+1} dxdy\), which presents challenges related to the integration of complex expressions involving logarithmic and arctangent functions.

Discussion Character

  • Exploratory, Conceptual clarification, Mathematical reasoning, Problem interpretation

Approaches and Questions Raised

  • Participants discuss various strategies for simplifying the integral, including the use of arctangent identities and the potential benefits of changing the order of integration. Questions arise about the appropriateness of looking up integral values and the implications of different integration approaches.

Discussion Status

The discussion includes attempts to clarify the integration process and explore alternative methods. Some participants suggest breaking the region into vertical strips as a potentially advantageous approach, while others reflect on the complexity of the expressions involved. There is a recognition of the need to reconsider limits when changing the order of integration.

Contextual Notes

Participants note the complexity of the integral and the challenges posed by the functions involved, particularly in relation to logarithmic and arctangent integrations. There is an emphasis on understanding the geometric interpretation of the integral in the xy-plane.

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Homework Statement



\int_{0}^{8} \int_{y^{1/3}}^{2} \frac{1}{x^4+1} dxdy

Homework Equations



Completing the square.

The Attempt at a Solution



This integral is disgusting. It literally took me 4 sheets of paper to do the partial fraction decomposition and then integrate the inner integral giving me an expression involving ln(compicated y expression) + ln(a number) + 1/2arctan(more garbage) + 1/2arctan(even more garbage).

Now I have an even more complicated integral to do and I have no clue how to approach it as I've never integrated arctan and integrating ln is nuts.

Any pointers on this one?
 
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Yeah I know I could easily look it up, but I don't think that's going to help.
 
The way I see it, your 1/2 arctan(garbage) expressions have two quite similar sorts of garbage, (x√2-1) and (x√2+1). How about using this?

arctan(a) + arctan(b) = arctan( (a+b)/(1-ab) )

That should lead you to arctan( (something with x) / (1 - something with x²) ). Partial fractions and then integrating two expressions like arctan(something) should be possible.

The ln expression I get looks like this:

ln ( (something with x² + something with x) / (something with x² - something with x) )

Try multiplying top and bottom of that fraction by the numerator to simplify the denominator. Maybe that'll help.
 
A good general rule of thumb is to think about what you are doing before you try to do it.

Bearing that in mind, what are you doing by computing that integral? Answer: integrating the function f(x)=\frac{1}{x^4+1} over some region in the xy-plane. Is the given form of the integral the only way to accomplish that? Is it the best way? What happens if you break the region down into vertical strips instead of horizontal strips and integrate over y first?
 
gabbagabbahey said:
What happens if you break the region down into vertical strips instead of horizontal strips and integrate over y first?

Unless I'm mistaken, something very nice happens then, and the double integral suddenly becomes

(some number) times ln (some other number)

Very clever. That's a strategy well worth remembering.
 
gabbagabbahey said:
A good general rule of thumb is to think about what you are doing before you try to do it.

Bearing that in mind, what are you doing by computing that integral? Answer: integrating the function f(x)=\frac{1}{x^4+1} over some region in the xy-plane. Is the given form of the integral the only way to accomplish that? Is it the best way? What happens if you break the region down into vertical strips instead of horizontal strips and integrate over y first?

Yes i completely forgot about changing my limits! Thank you very much, I solved it :).
 

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