Page 1 of 2 12 LastLast
Results 1 to 15 of 16

Math Help - Complicated Integration

  1. #1
    Newbie
    Joined
    Dec 2007
    Posts
    6

    Complicated Integration

    Is there anybody here who know hows to solve any of the problems below? I've tried some methods but none of then worked.

    <br />
\int \frac{cos(x)}{x^2+a^2}\,dx<br />

    <br />
\int \frac{xsin(x)}{x^2+a^2}\,dx<br />


    <br />
\int \frac{x^\frac{1}{3}}{x^2+1}\,dx<br />


    <br />
 \int \frac{log(x)}{x^2+1}\,dx<br />

    It would be really helpfull if anybody knows any of the problems above.
    Thanks for the help
    Last edited by brenorocha; December 2nd 2007 at 02:49 AM.
    Follow Math Help Forum on Facebook and Google+

  2. #2
    Global Moderator

    Joined
    Nov 2005
    From
    New York City
    Posts
    10,616
    Thanks
    9
    Maybe you have the limit -\infty to \infty in that case some complex analysis will do it.
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Newbie
    Joined
    Dec 2007
    Posts
    6

    Ops...

    I forgot to mention that, the limits are 0 and \infty . Isn't there a way of solving that without putting complex numbers involved?
    I mean, I haven't learnt it yet, anyway, could you please post the solution of any of then with that complex analysis
    Follow Math Help Forum on Facebook and Google+

  4. #4
    Math Engineering Student
    Krizalid's Avatar
    Joined
    Mar 2007
    From
    Santiago, Chile
    Posts
    3,654
    Thanks
    13
    Quote Originally Posted by brenorocha View Post
    <br />
\int_0^\infty \frac{log(x)}{x^2+1}\,dx<br />
    This one is fairly known. Substitute u=\ln x.

    You'll get an odd function on the symmetric interval (-\infty,\infty), so the expected answer is zero.
    Follow Math Help Forum on Facebook and Google+

  5. #5
    Global Moderator

    Joined
    Nov 2005
    From
    New York City
    Posts
    10,616
    Thanks
    9
    Quote Originally Posted by brenorocha View Post
    <br />
\int \frac{cos(x)}{x^2+a^2}\,dx<br />
    \int_{-\infty}^{\infty} \frac{\cos x}{x^2+a^2} dx where a>0.
    Define,
    f(z) = \frac{e^{iz}}{z^2+a^2}
    And integrate over a large semi-circle with radius R>0 with R\to \infty.
    As a result we will have,
    \int_{-\infty}^{\infty} \frac{e^{ix}}{x^2+a^2}dx = 2\pi i \left( \mbox{res}(f,ai) \right)
    Thus,
    \int_{-\infty}^{\infty} \frac{\cos x}{x^2+a^2}dx + i \int_{-\infty}^{\infty}\frac{\sin x}{x^2+a^2}dx = 2\pi i \left( \frac{e^{i^2a}}{ai+ai}\right) = 2\pi i \left( \frac{e^{-a}}{2ai} \right)= \frac{\pi}{ae^a}
    Follow Math Help Forum on Facebook and Google+

  6. #6
    Newbie
    Joined
    Dec 2007
    Posts
    6

    Thanks

    Thank you for the help...
    I'll try to solve the others now.
    Just to notice, how do I calculate the res() that is in your solution?
    The other ( \frac{1}{3}\int_0^{\infty}\frac{1}{t(t^6+1)} dt ), how can I solve it? Trigonometric substitution? Parts? or Partial fraction? which is the easiest?
    Follow Math Help Forum on Facebook and Google+

  7. #7
    Super Member PaulRS's Avatar
    Joined
    Oct 2007
    Posts
    571
    Quote Originally Posted by brenorocha View Post
    The other ( \frac{1}{3}\int_0^{\infty}\frac{1}{t(t^6+1)} dt ), how can I solve it? Trigonometric substitution? Parts? or Partial fraction? which is the easiest?
    I'd let u=t^6

    But you don't get that integral with the subsitution x^{1/3}=t

    \int{\frac{x^{1/3}}{x^2+1}}dx=3\cdot{\int{\frac{t^3}{t^6+1}}}dt
    Last edited by PaulRS; December 2nd 2007 at 10:02 AM.
    Follow Math Help Forum on Facebook and Google+

  8. #8
    Math Engineering Student
    Krizalid's Avatar
    Joined
    Mar 2007
    From
    Santiago, Chile
    Posts
    3,654
    Thanks
    13
    \varphi=\int_0^\infty {\frac{{x^{1/3} }}<br />
{{x^2 + 1}}\,dx} .

    As said, make the substitution x=u^3,

    \varphi = \int_0^\infty {\frac{{x^{1/3} }}<br />
{{x^2 + 1}}\,dx} = \int_0^\infty {\frac{{3u^3 }}<br />
{{u^6 + 1}}\,du} .

    Define another substitution u=\frac1v,

    \varphi = \int_0^\infty {\frac{{3v}}<br />
{{v^6 + 1}}\,dv} .

    By symmetry,

    \varphi = \int_0^\infty {\frac{{3x^3 }}<br />
{{x^6 + 1}}\,dx} = \int_0^\infty {\frac{{3x}}<br />
{{x^6 + 1}}\,dx} .

    So, 2\varphi = 3\int_0^\infty {\frac{{x\left( {x^2 + 1} \right)}}<br />
{{\left( {x^2 + 1} \right)\left( {x^4 - x^2 + 1} \right)}}\,dx} = 3\int_0^\infty {\frac{x}<br />
{{x^4 - x^2 + 1}}} .

    Final substitution \mu=x^2,

    2\varphi = \frac{3}<br />
{2}\int_0^\infty {\frac{{d\mu }}<br />
{{\mu ^2 - \mu + 1}}} = \frac{3}<br />
{2} \cdot \frac{{4\pi }}<br />
{{3\sqrt 3 }}.

    And we happily get \varphi = \frac{\pi }<br />
{{\sqrt 3 }}.

    Follow Math Help Forum on Facebook and Google+

  9. #9
    Forum Admin topsquark's Avatar
    Joined
    Jan 2006
    From
    Wellsville, NY
    Posts
    9,854
    Thanks
    321
    Awards
    1
    Quote Originally Posted by Krizalid View Post
    By symmetry,

    \varphi = \int_0^\infty {\frac{{3x^3 }}<br />
{{x^6 + 1}}\,dx} = \int_0^\infty {\frac{{3x}}<br />
{{x^6 + 1}}\,dx} .
    What is the symmetry argument here? (I'm going to kick myself for not spotting the answer, I just know it! )

    -Dan
    Follow Math Help Forum on Facebook and Google+

  10. #10
    Newbie
    Joined
    Dec 2007
    Posts
    6
    Quote Originally Posted by Krizalid View Post
    Define another substitution u=\frac1v,

    \varphi = \int_0^\infty {\frac{{3v}}<br />
{{v^6 + 1}}\,dv} .

    Isn't there a signal missing?
    \varphi = \int_0^\infty {\frac{{-3v}}<br />
{{v^6 + 1}}\,dv} .
    Follow Math Help Forum on Facebook and Google+

  11. #11
    Forum Admin topsquark's Avatar
    Joined
    Jan 2006
    From
    Wellsville, NY
    Posts
    9,854
    Thanks
    321
    Awards
    1
    Quote Originally Posted by brenorocha View Post
    Isn't there a signal missing?
    \varphi = \int_0^\infty {\frac{{-3v}}<br />
{{v^6 + 1}}\,dv} .
    No. The integral after the substitution comes out to be
    \int_{\infty}^0 \frac{ \frac{1}{v} }{ \frac{1}{v^6} + 1 } \cdot -\frac{dv}{v^2}

    The negative sign restores the integral to a \int_0^{\infty} type.

    -Dan
    Follow Math Help Forum on Facebook and Google+

  12. #12
    Newbie
    Joined
    Dec 2007
    Posts
    6
    Quote Originally Posted by Krizalid View Post
    \int_0^\infty \frac{log(x)}{x^2+1}\,dx
    Quote Originally Posted by Krizalid View Post

    This one is fairly known. Substitute u=\ln x.

    You'll get an odd function on the symmetric interval (-\infty,\infty), so the expected answer is zero.
    Making that substitution I got:

     \int_0^\infty \frac{u e^u}{e^{2u}+1}\,du

    But I can't see why it would be 0. Is that really that obvious????
    Follow Math Help Forum on Facebook and Google+

  13. #13
    Math Engineering Student
    Krizalid's Avatar
    Joined
    Mar 2007
    From
    Santiago, Chile
    Posts
    3,654
    Thanks
    13
    Quote Originally Posted by topsquark View Post
    What is the symmetry argument here?
    It's just turnin' back u,v into x. There's no difference between the integrals

    \int_0^\infty {\frac{{3u^3 }}<br />
{{u^6 + 1}}\,du} = \int_0^\infty {\frac{{3x^3 }}<br />
{{x^6 + 1}}\,dx} .

    The value of the integral is always the same.

    Quote Originally Posted by brenorocha View Post

    Making that substitution I got:

     \int_0^\infty \frac{u e^u}{e^{2u}+1}\,du
    Make again the substitution and check the lower integration limit.
    Follow Math Help Forum on Facebook and Google+

  14. #14
    Newbie
    Joined
    Dec 2007
    Posts
    6
    That's right... I forgot to change the limits... sorry
    Thanks for the help
    there is just one left now...

    \int \frac{xsin(x)}{x^2+a^2}\,dx
    Follow Math Help Forum on Facebook and Google+

  15. #15
    Forum Admin topsquark's Avatar
    Joined
    Jan 2006
    From
    Wellsville, NY
    Posts
    9,854
    Thanks
    321
    Awards
    1
    Quote Originally Posted by Krizalid View Post
    It's just turnin' back u,v into x. There's no difference between the integrals

    \int_0^\infty {\frac{{3u^3 }}<br />
{{u^6 + 1}}\,du} = \int_0^\infty {\frac{{3x^3 }}<br />
{{x^6 + 1}}\,dx} .

    The value of the integral is always the same.
    Like I said:

    Thanks.

    -Dan
    Follow Math Help Forum on Facebook and Google+

Page 1 of 2 12 LastLast

Similar Math Help Forum Discussions

  1. Complicated Integral
    Posted in the Calculus Forum
    Replies: 3
    Last Post: February 20th 2010, 08:29 AM
  2. the equations may seem more complicated?
    Posted in the Pre-Calculus Forum
    Replies: 1
    Last Post: November 25th 2009, 02:26 PM
  3. Integration complicated problems
    Posted in the Calculus Forum
    Replies: 1
    Last Post: March 20th 2009, 07:59 PM
  4. Complicated Series
    Posted in the Calculus Forum
    Replies: 1
    Last Post: May 16th 2008, 07:18 PM
  5. complicated integral
    Posted in the Calculus Forum
    Replies: 7
    Last Post: April 22nd 2008, 04:45 PM

Search Tags


/mathhelpforum @mathhelpforum