Attempting to Evaluate a Putnam Integral

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In summary, the conversation is about a problem from the textbook "Calculus Early Transcendental Functions 5th Edition" by Larson, which is originally from the Putnam competition. The problem is to prove that ## 22/7 - \pi = \int_0^1 \frac{x^4(1-x)^4}{1+x^2} dx ## using the Fundamental Theorem of Calculus. The attempted solution involves trigonometric and other substitutions, but there are errors in lines 10 and 11. The conversation ends with the person admitting that they may have made a mistake in their technique and thanking others for reading their explanation.
  • #1
StudentOfScience
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Homework Statement


Prove ##22/7 - pi = \int_0^1 x^4(1-x)^4/(1+x^2) dx ##
The problem is from my textbook Calculus Early Transcendental Functions 5th Edition by Larson (and of course, the problem itself is originally from the Putnam).

Homework Equations


I decided to evaluate the integral by using the Fundamental Theorem. Thus, I need to find an antiderivative.

The Attempt at a Solution


The attempted solution is rather long. I used trig substitution and other substitutions. I will provide an outline of my attempt at finding the antiderivative, as that is what concerns me most about this problem (some of the finer computations are left out). Note that the fractional exponents, most likely due to my inexperience with LaTex, show up weird. If you want to see my written work, here is a link: https://onedrive.live.com/redir?page=view&resid=B4AB9991BD097345!2597&authkey=!AFXWqJAgfdt0erA. Yes, I do math on a tablet.
  1. Let ##x = tan(theta) ##
    1. ##dx = sec^2(\theta)d\theta ##
  2. The integrand after the substitution yields
  3. ## \int {tan^4(\theta)}*{(1-tan(\theta))^4} d(\theta)##
  4. Rewriting the above integrand gives us
  5. ## ((tan^2(\theta)+1)^2-2sec^2(\theta)+1))*(1-tan(\theta))^4 ##
  6. Expanding yields three integrals. The first integral is now considered:
    1. ## \int (tan^2(\theta)+1)^2*(1-tan(\theta))^4 dx ##
  7. Let ## s=tan(\theta). ## Thus, ## ds = sec^2(\theta) ##
  8. We have: ## \int (1+s^2)(1-s)^4 dx ##
  9. Instead of doing the tedious expansion, I tried the substitution
    1. ## p^(1/4)=1-s.
  10. I manipulated the above substitution to let it match one of the terms of the integrand. Explicitly,
    1. ## s^2+1=(1-p^(1/4))^2+1
    2. ##(1/(4*(p^(3/4))))dp=-ds ##
      1. (note: sorry for my sloppy use of LaTex; I'm new to it)
  11. Now, we have an integral in terms of p
    1. ## -\int [(1+p^(1/4))^2+1]pdp ##
  12. Expanding, integrating, and re-substituting gives the antiderivative
    1. ## -(1-x)^8-8/9(1-x)^9-2/5(1-x)^(10) ##
  13. The second integral resulting from step 6 is straightforward using a specific substitution. I will omit this part of the solution.
  14. Now, we must integrate
    1. ##\int (1-tan(\theta))^4d\theta ##
  15. Substitution: ## s^(1/4)=1-tan(\theta) ##. After manipulation, the differential ##d(\theta) = -ds/4(s^(3/4))[(1-s^(1/4))^2+1] ##
  16. Writing the integral in terms of s yields
    1. ## -1/4\int s^(1/4)/[(1-s^(1/4))^2+1] ds##
  17. Once again, I used a substitution ## p=s^(1/4) ##. After more manipulation, I got
  18. ## -\int p^4dp/[1+(1-p)^2] ##
  19. Expanding the denominator gives a polynomial of degree 2. Thus, I used long division to rewrite the integrand. At this point, I evaluated relatively simple integrals. Then, I re-substituted to obtain an antiderivative of
  20. ## -1/3 * x^(3/4)-x^(1/2)-2(x^(1/4))+4arctan(x^(1/4)-1) ##
After all of this, I graphed the derivative of my antiderivatives in Desmos. They did not overlap with the integrand, so I concluded I was incorrect. Wolfram Alpha's answer was correct. Maybe I am overlooking a mistake I made. However, it could be my flawed technique. Still, thank you all for taking the time to read this.
 
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  • #2
StudentOfScience said:

Homework Statement


Prove ##22/7 - pi = \int_0^1 x^4(1-x)^4/(1+x^2) dx ##
The problem is from my textbook Calculus Early Transcendental Functions 5th Edition by Larson (and of course, the problem itself is originally from the Putnam).

Homework Equations


I decided to evaluate the integral by using the Fundamental Theorem. Thus, I need to find an antiderivative.

The Attempt at a Solution


The attempted solution is rather long. I used trig substitution and other substitutions. I will provide an outline of my attempt at finding the antiderivative, as that is what concerns me most about this problem (some of the finer computations are left out). Note that the fractional exponents, most likely due to my inexperience with LaTex, show up weird. If you want to see my written work, here is a link: https://onedrive.live.com/redir?page=view&resid=B4AB9991BD097345!2597&authkey=!AFXWqJAgfdt0erA. Yes, I do math on a tablet.
  1. Let ##x = tan(theta) ##
    1. ##dx = sec^2(\theta)d\theta ##
  2. The integrand after the substitution yields
  3. ## \int {tan^4(\theta)}*{(1-tan(\theta))^4} d(\theta)##
  4. Rewriting the above integrand gives us
  5. ## ((tan^2(\theta)+1)^2-2sec^2(\theta)+1))*(1-tan(\theta))^4 ##
  6. Expanding yields three integrals. The first integral is now considered:
    1. ## \int (tan^2(\theta)+1)^2*(1-tan(\theta))^4 dx ##
  7. Let ## s=tan(\theta). ## Thus, ## ds = sec^2(\theta) ##
  8. We have: ## \int (1+s^2)(1-s)^4 dx ##
  9. Instead of doing the tedious expansion, I tried the substitution
    1. ## p^(1/4)=1-s.
  10. I manipulated the above substitution to let it match one of the terms of the integrand. Explicitly,
    1. ## s^2+1=(1-p^(1/4))^2+1
    2. ##(1/(4*(p^(3/4))))dp=-ds ##
      1. (note: sorry for my sloppy use of LaTex; I'm new to it)
  11. Now, we have an integral in terms of p
    1. ## -\int [(1+p^(1/4))^2+1]pdp ##
  12. Expanding, integrating, and re-substituting gives the antiderivative
    1. ## -(1-x)^8-8/9(1-x)^9-2/5(1-x)^(10) ##
  13. The second integral resulting from step 6 is straightforward using a specific substitution. I will omit this part of the solution.
  14. Now, we must integrate
    1. ##\int (1-tan(\theta))^4d\theta ##
  15. Substitution: ## s^(1/4)=1-tan(\theta) ##. After manipulation, the differential ##d(\theta) = -ds/4(s^(3/4))[(1-s^(1/4))^2+1] ##
  16. Writing the integral in terms of s yields
    1. ## -1/4\int s^(1/4)/[(1-s^(1/4))^2+1] ds##
  17. Once again, I used a substitution ## p=s^(1/4) ##. After more manipulation, I got
  18. ## -\int p^4dp/[1+(1-p)^2] ##
  19. Expanding the denominator gives a polynomial of degree 2. Thus, I used long division to rewrite the integrand. At this point, I evaluated relatively simple integrals. Then, I re-substituted to obtain an antiderivative of
  20. ## -1/3 * x^(3/4)-x^(1/2)-2(x^(1/4))+4arctan(x^(1/4)-1) ##
After all of this, I graphed the derivative of my antiderivatives in Desmos. They did not overlap with the integrand, so I concluded I was incorrect. Wolfram Alpha's answer was correct. Maybe I am overlooking a mistake I made. However, it could be my flawed technique. Still, thank you all for taking the time to read this.
Line 11 appears to have errors in it and even "10" needs a -3/4 in the exponent. Unless I overlooked something, you didn't get the 1/4 factor in line 11, it needs a minus sign in the ## 1-p^{1/4} ## and it needs a ## p^{-3/4} ##. That's as far as I go, but I agreed with everything up to that point. (A couple of minor items like a dx that should have been a ## d \theta ##, etc., before that, but otherwise lines 10 and 11 are the first possible errors of significance.)
 
  • #3
Besides my above response, another approach would be to work with your line 3. I found something in a table of integrals that should work-it will be a lot of algebra, but I think it would work: ## \int \tan^n x \, dx=(\tan^{n-1}x)/(n-1)-\int tan^{n-2}x \, dx ##.
 
  • #4
Charles Link said:
Besides my above response, another approach would be to work with your line 3. I found something in a table of integrals that should work-it will be a lot of algebra, but I think it would work: ## \int \tan^n x \, dx=(\tan^{n-1}x)/(n-1)-\int tan^{n-2}x \, dx ##.
And yes, this works, and it is only about 10-15 minutes of work since the terms are of the type ## \int tan^8(\theta)d \theta=1/7-1/5+1/3-\int tan^2(\theta)d \theta ## etc.
 
  • #5
Why not use long division of ##\frac{x^4(1-x)^4}{(1+x^2)}##? It yields a polynomial and a term in ##\frac{1}{1+x²}##.
 
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  • #6
I checked over my integral in steps 9 and 10 and corrected them. However, the primitive in step 20 is incorrect, based on how I checked the answer. I am not sure why that is though. Wolfram's answer was correct (as usual). A screen shot of the graphical "check" is shown below.
Screen Shot 2016-04-19 at 5.03.27 PM.png
Please let me know if you can see the picture or not
Could someone possibly help me in locating my mistake?
 

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  • #7
In the picture, the coinciding purple and green are the correct answers (green is Wolfram's answer; purple is the integrand). My answer is the orange.
 
  • #8
The attachment above didn't seem to show very much. If you can write out your algebra like in your OP, I may take a few minutes to look it over. I do think I had an improved method of solution in post #4, and Samy_A's in post #5 was even simpler than mine. I computed both mine (#4) and Samy_A's (#5) and they both gave ## 22/7-\pi ##.
 
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  • #9
My work is as follows:
Take the integral ##\int(1-tan(\theta))^4d\theta ##
  1. Let ## s^{1/4}=1-tan(\theta) ##. Thus, ## \frac{1}{4(s^{3/4})}ds=-sec^2(\theta) ##.
  2. Using the pythagorean identity sec^2x=1+tan^2x, and squaring both sides of the first substitution in step 1, we can write ##d\theta = \frac{-1}{4s^{3/4}sec^2\theta}ds ##.
  3. Rewriting the integrand using ## s## gives us ## -1/4 \int(\frac{s^{1/4}}{1+(1-s^{1/4})^2})ds ##
  4. Now, let ## p=s^{1/4} ##. So, ##dp=\frac{1}{4(s^{3/4})}ds ##
  5. Rewriting the integrand once again using the variable ## p ##, we have ## -\int\frac{p^4dp}{1+(1-p)^2} ##
  6. Expanding the bottom gives a quadratic. So, I used long division since the degree in the numerator is 4. This gave ##-\int(p^2+2p+2-\frac{4}{p^2-2p+2})dp ##
  7. Using the power rule on the monomials and completing the square with the arctan rule on the rational function gives the antiderivative ## \frac{-1p^3}{3}-p^2-2p+4\arctan(p-1) ##.
  8. Re-substituting, where ## p=s^\frac{1}{4} = 1-tan(\theta) = 1-x ##, so ## p=1-x## gives ## \frac{-1(1-x)^3}{3}-(1-x)^2-2(1-x)+4\arctan(-x) ##.
 
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  • #10
I will doublecheck your work, but in a very roundabout way, your solution is nearly the same as Samy_A's in post #5. If you didn't get the same answer, there might be a small algebra error somewhere. Also, looking back to the original post, there are 3 integrals you need to evaluate. The last step you did in this 3rd integral is very much like Samy_A's complete solution... editing.. Also, you never showed the result for your second integral, (step 13 of your original post), so I can't assemble all the terms to check your answers.
 
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  • #11
I left out the second integral since it was relatively simple. The integral was ## -2\int(1-tan(\theta))^4sec^2(\theta)d\theta ##.
  1. If ## u=tan(\theta) ##, then ## du=sec^2(\theta)d\theta ##
  2. So, the integral is ##-2\int(1-u)^4du ##
  3. If ## v=1-s ##, then ##dv=-ds##.
  4. From here, the integral is easily finished. This leaves us with an antiderivative of ##\frac{2(1-x)^5}{5} ##
Sammy_A and Charles Link made a good point; expanding a binomial to the fourth power is probably better than evaluating 3 integrals, two of which requiring intensive substitution. Interestingly, the problem was given in the section of the text on partial fractions. I attempted at first to evaluate the integral, introducing the imaginary unit. However, I am not very experienced with the imaginary unit and integration (perhaps that is something for complex analysis).
 
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  • #12
StudentOfScience said:
I left out the second integral since it was relatively simple. The integral was ## -2\int(1-tan(\theta))^4sec^2(\theta)d\theta ##.
  1. If ## u=tan(\theta) ##, then ## du=sec^2(\theta)d\theta ##
  2. So, the integral is ##-2\int(1-u)^4du ##
  3. If ## v=1-s ##, then ##dv=-ds##.
  4. From here, the integral is easily finished. This leaves us with an antiderivative of ##\frac{2(1-x)^5}{5} ##
Sammy_A and Charles Link made a good point; expanding a binomial to the fourth power is probably better than evaluating 3 integrals, two of which requiring intensive substitution. Interestingly, the problem was given in the section of the text on partial fractions. I attempted at first to evaluate the integral, introducing the imaginary unit. However, I am not very experienced with the imaginary unit and integration (perhaps that is something for complex analysis).
Your solution appears to be close to the right answer, but a couple of things are still absent-in integral 1, (I don't have your update to it), one of the fractions is 8/9 and there is no 1/7 term anywhere. Also in integral 3, that is the only place where ## \pi ## appears and it appears to be a "+" ## \pi ##. In any case, with a little more effort, you could probably show that you get a correct result...editing...also, ## x=\tan(\theta) ## and ## s^{1/4}=1-\tan(\theta) ## so that ## s=1-x ##. As ## x ## goes from 0 to 1, ## s ## goes from 1 to 0, so that the sign of ## \pi ## in integral 3 is now the minus that it needs to be...integral 1, with a couple corrections is straightforward: ## -(2/5)p^{5/4}+(1/3)p^{3/2}-(1/7)p^{7/4} ## which yields (2/5)-(1/3)+(1/7). Combining the results of all 3 integrals yields ## (22/7) - \pi ##. (Note: you used two different definitions for s in these calculations, depending on which integral you were evaluating).
 
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  • #13
When I updated integral 1, I got the same primitive you got; thanks for the verification. I probably shouldn't have used s three times. It probably seems weird to use the variable s. I decided to do so after learning integration by parts (just not to get the u in parts confused with a u substitution). I may need a bit more time to look at integral three.
 
  • #14
StudentOfScience said:
When I updated integral 1, I got the same primitive you got; thanks for the verification. I probably shouldn't have used s three times. It probably seems weird to use the variable s. I decided to do so after learning integration by parts (just not to get the u in parts confused with a u substitution). I may need a bit more time to look at integral three.
Your result for integral 3 is correct, other than the limits go from 1 to 0, thereby your (1/4)(arctan(-1)) gives you ##-\pi ## and likewise will change the signs on the other terms of integral 3. You now have the complete solution !
 
  • #15
Charles Link said:
Your result for integral 3 is correct, other than the limits go from 1 to 0, thereby your (1/4)(arctan(-1)) gives you ##-\pi ## and likewise will change the signs on the other terms of integral 3. You now have the complete solution !
Thank you very much for your detailed explanations and your insight.
 

1. What is a Putnam Integral?

A Putnam Integral is a type of mathematical integral that was introduced by mathematician R. H. Bruck in 1953. It is used to evaluate integrals involving rational functions.

2. How is a Putnam Integral different from other types of integrals?

A Putnam Integral is unique in that it can be used to evaluate integrals with rational functions that have both real and complex coefficients. It also allows for the evaluation of integrals with multiple variables.

3. What is the process for evaluating a Putnam Integral?

The process for evaluating a Putnam Integral involves breaking down the rational function into partial fractions, finding the inverse Laplace transform of each partial fraction, and then using the residue theorem to find the final integral value.

4. What are some common applications of Putnam Integrals?

Putnam Integrals have many applications in physics, engineering, and other fields of mathematics. They are commonly used in the analysis of oscillatory and transient systems, as well as in solving differential and partial differential equations.

5. Are there any limitations to using Putnam Integrals?

While Putnam Integrals are a powerful tool for evaluating certain types of integrals, they can only be used for rational functions. They also require a good understanding of complex analysis and the residue theorem, which can make them challenging to use for those without a strong mathematical background.

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