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Math Help - Collecting integral problems

  1. #1
    Math Engineering Student
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    Collecting integral problems

    Yes, post an interesting integral or a problem related to integration (like a sum, double sum, etc which can be computed applyin' integration), 'cause I want to collect problems and their solutions to make a PDF.

    I'm not gonna start yet, may be in few hours, but if someone wants to start, no problem.

    (I hope we can share lots of interesting integrals.)
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  2. #2
    Senior Member polymerase's Avatar
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    \int \frac {sin\;xcos\;x}{sin^4x+cos^4x}\;dx

    too easy? too boring?
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  3. #3
    Senior Member polymerase's Avatar
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    can i get a copy of the PDF when your done?
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    Quote Originally Posted by polymerase View Post
    \int \frac {sin\;xcos\;x}{sin^4x+cos^4x}\;dx

    too easy? too boring?
    Cool. . (Set t=\sin^2 x ).
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  5. #5
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    Quote Originally Posted by polymerase View Post
    \int \frac {sin\;xcos\;x}{sin^4x+cos^4x}\;dx

    too easy? too boring?
    Sorry to say, but too easy and too boring. (Read above, it's about interesting integrals.)

    Quote Originally Posted by polymerase View Post
    can i get a copy of the PDF when your done?
    Yes. I hope people post problems.

    ---

    Okay, here's my problem: Evaluate \int_0^1 {\int_0^1 {\frac{{x - 1}}<br />
{{(1 + xy)\ln (xy)}}\,dx} \,dy} .

    Solution:

    For the inner integral, substitute u=xy. After reversing integration order we have

    \int_0^1 {\int_0^1 {\frac{{x - 1}}<br />
{{(1 + xy)\ln (xy)}}\,dx} \,dy}= \underbrace {\int_0^1 {\frac{{1 - u}}<br />
{{(1 + u)\ln u}}\,du} }_\ell  + \ln 2.

    Now the remaining challenge is to compute \ell. Expanding as a sum we get

    \ell  = \sum\limits_{k = 0}^\infty  {( - 1)^k \ln \frac{{k + 1}}<br />
{{k + 2}}}  = \ln \frac{{1 \cdot 3 \cdot 3 \cdot 5 \cdot  \cdots  \cdot (2n - 1)(2n + 1)}}<br />
{{2 \cdot 2 \cdot 4 \cdot 4 \cdot  \cdots  \cdot (2n)(2n)}}.

    Inside logarithm we have the Wallis Product and its limit is \frac2\pi. The rest follows.
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    If x>0 show that: \int_0^{\infty} \frac{\cos (xt) + t \sin (xt)}{1+t^2} dt = \pi e^{-x}.
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  7. #7
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    Hey K:

    Here's one you're probably familiar with, but if not then it's a good one:

    \int_{-\infty}^{\infty}\frac{1}{\sqrt{(x^{2}+a^{2})(x^{2}  +b^{2})}}dx=\frac{\pi}{agm(a,b)}

    I think this was one of Gauss's babies.
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  8. #8
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    Okay guys, but I dunno if you read well above, you need to post problem + solution. (To add those problems into a PDF.)

    Here's another problem: Evaluate \sum\limits_{k = 0}^\infty  {\frac{1}<br />
{{(k + 2)k!}}} .

    Start with series expansion for e^x and multiply by x,

    e^x  = \sum\limits_{k = 0}^\infty  {\frac{{x^k }}<br />
{{k!}}}  \implies xe^x  = \sum\limits_{k = 0}^\infty  {\frac{{x^{k + 1} }}<br />
{{k!}}} .

    Integrate \int_0^x {ue^u \,du}  = \sum\limits_{k = 0}^\infty  {\frac{{x^{k + 2} }}<br />
{{(k + 2)k!}}} .

    The integral can be evaluated easily via integration by parts, this yields

    xe^x  - e^x  + 1 = \sum\limits_{k = 0}^\infty  {\frac{{x^{k + 2} }}<br />
{{(k + 2)k!}}} . By setting x=1 we have the desired answer.
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  9. #9
    Senior Member polymerase's Avatar
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    Is this interesting by your standards?

    \lim_{x\to\infty}\left(\frac{n}{1+3n^2}+\frac{n}{4  +3n^2}+\frac{n}{9+3n^2}+...+\frac{n}{n^2+3n^2}\rig  ht)=?

    Ill post the answer if it is.
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  10. #10
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    Quote Originally Posted by polymerase View Post
    \lim_{n\to\infty}\left(\frac{n}{1+3n^2}+\frac{n}{4  +3n^2}+\frac{n}{9+3n^2}+...+\frac{n}{n^2+3n^2}\rig  ht)=?
    \sum_{k=1}^n \frac{n}{k^2+3n^2}  = \sum_{k=1}^n \frac{1}{\left( \frac{k}{n} \right)^2 + 3} \cdot \frac{1}{n}

    This is the Riemann sum for \int_0^1 \frac{dx}{x^2+3} with equal sub-intervals of length 1/n using right-endpoints.

    Thus, (by fundamental theorem) this limit converges to \sqrt{3} \tan^{-1} \frac{x}{\sqrt{3}} \bigg|_0^1 = \sqrt{3} \tan^{-1} \frac{1}{\sqrt{3}} = \frac{\pi \sqrt{3}}{6}
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  11. #11
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    Yeah, we had an obviously Riemann sum there.

    Riemann sums are routine problems, unless you to post some interesting limit.
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  12. #12
    Senior Member polymerase's Avatar
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    Quote Originally Posted by ThePerfectHacker View Post
    \sum_{k=1}^n \frac{n}{k^2+3n^2}  = \sum_{k=1}^n \frac{1}{\left( \frac{k}{n} \right)^2 + 3} \cdot \frac{1}{n}

    This is the Riemann sum for \int_0^1 \frac{dx}{x^2+3} with equal sub-intervals of length 1/n using right-endpoints.

    Thus, (by fundamental theorem) this limit converges to \sqrt{3} \tan^{-1} \frac{x}{\sqrt{3}} \bigg|_0^1 = \sqrt{3} \tan^{-1} \frac{1}{\sqrt{3}} = \frac{\pi \sqrt{3}}{6}
    One tiny problem, it converges to \frac{1}{\sqrt{3}}arctan\left(\frac{x}{\sqrt{3}}\r  ight) Answer should be \frac{\pi}{6\sqrt{3}}
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  13. #13
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    Quote Originally Posted by Krizalid View Post
    Yes, post an interesting integral or a problem related to integration (like a sum, double sum, etc which can be computed applyin' integration), 'cause I want to collect problems and their solutions to make a PDF.

    I'm not gonna start yet, may be in few hours, but if someone wants to start, no problem.

    (I hope we can share lots of interesting integrals.)
    Well, this one is probably more beautiful than interesting. You've no doubt met it. If it nevertheless meets your criteria for interesting, I'll take the time to post a solution.

    \int_0^\infty \frac{u^{x-1}}{e^{u} - 1} \, du = \zeta(x) \Gamma(x),

    where \zeta (x) and \Gamma (x) are, respectively, the famous zeta and Gamma functions.
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  14. #14
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    Quote Originally Posted by Krizalid View Post
    Yes, post an interesting integral or a problem related to integration (like a sum, double sum, etc which can be computed applyin' integration), 'cause I want to collect problems and their solutions to make a PDF.

    I'm not gonna start yet, may be in few hours, but if someone wants to start, no problem.

    (I hope we can share lots of interesting integrals.)
    Ditto to my previous post:

    \int_0^1 \frac{1}{x^x} \, dx = \frac{1}{1^1} + \frac{1}{2^2} + \frac{1}{3^3} + ....
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  15. #15
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    Quote Originally Posted by mr fantastic View Post
    Well, this one is probably more beautiful than interesting. You've no doubt met it. If it nevertheless meets your criteria for interesting, I'll take the time to post a solution.

    \int_0^\infty \frac{u^{x-1}}{e^{u} - 1} \, du = \zeta(x) \Gamma(x),

    where \zeta (x) and \Gamma (x) are, respectively, the famous zeta and Gamma functions.
    That is nice . But that is not so hard to prove. It should be worthy to be such a problem.

    This is Mine 86th Post!!!
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