Page 1 of 6 12345 ... LastLast
Results 1 to 15 of 90

Math Help - Integrals

  1. #1
    MHF Contributor Mathstud28's Avatar
    Joined
    Mar 2008
    From
    Pennsylvania
    Posts
    3,641

    Integrals

    Hey mathstud, you know what might be fun?. Go to the link I posted above and actually derive the solutions they give for the integrals. You have probably done that already.

    Some of them are probably very challenging.
    You know me well sir. I already was there deriving them, now most of them I can see what to do. But one is alluding me . Now I haven't given it much time but

    \int_0^{\infty}\frac{e^{-ax}-e^{-bx}}{x\csc(px)}

    Ok my first move was to do the obvious double integral

    \frac{e^{-ax}-e^{-bx}}{x}=\int_a^{b}e^{-yx}dy

    So we have

    \int_0^{\infty}\csc(px)\int_a^{b}e^{-yx}~dy~dx=\int_0^{\infty}\int_a^{b}e^{-yx}\csc(px)~dy~dx

    Now rewriting this as (this is so people who aren't super accustomed to double integrals can see what I am doing)

    \iint_{R}e^{-yx}\csc(px)~dA

    Where R=\left\{(x,y):0\leq{x}\leq\infty~{a\leq{y}\leq{b}  }\right\}

    From here it is obvious that this region is a rectangle, so Fubini's Theorem applies

    So

    \int_0^{\infty}\int_a^{b}e^{-yx}\csc(px)~dy~dx=\int_a^{b}\int_0^{\infty}e^{-yx}\csc(px)~dx~dy

    Now this inner integral is bothering me, I cannot seem to get it in the eight minutes I have tried

    parts-no deal

    Letting J(\theta)=\int_0^{\infty}e^{-yx}\csc(px)e^{-\theta\sin(px)}dx-no go

    Any suggestions?


    I also tried some crazy thing where I let

    J(\theta)=\int_0^{\infty}\frac{e^{-xy\cos(px\theta)}}{\sin(px)}~dx

    And found that the integral I want is a constant difference from J'(0)
    Follow Math Help Forum on Facebook and Google+

  2. #2
    MHF Contributor
    Joined
    Apr 2008
    Posts
    1,092
    How are you getting from

    \int_0^{\infty}\frac{e^{-ax}-e^{-bx}}{x\csc(px)}

    to

    \int_0^{\infty}\csc(px)\int_a^{b}e^{-yx}~dy~dx=\int_0^{\infty}\int_a^{b}e^{-yx}\csc(px)~dy~dx?

    Shouldn't it be

    \int_0^{\infty}\int_a^{b}e^{-yx}\sin(px)~dy~dx?

    No suggestions on how to proceed with this, but \frac{1}{\csc x} = \sin x
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Rhymes with Orange Chris L T521's Avatar
    Joined
    May 2008
    From
    Santa Cruz, CA
    Posts
    2,844
    Thanks
    3
    Quote Originally Posted by Mathstud28 View Post
    You know me well sir. I already was there deriving them, now most of them I can see what to do. But one is alluding me . Now I haven't given it much time but

    \int_0^{\infty}\frac{e^{-ax}-e^{-bx}}{x\csc(px)}

    Ok my first move was to do the obvious double integral

    \frac{e^{-ax}-e^{-bx}}{x}=\int_a^{b}e^{-yx}dy

    So we have

    \color{red}\int_0^{\infty}\csc(px)\int_a^{b}e^{-yx}~dy~dx=\int_0^{\infty}\int_a^{b}e^{-yx}\csc(px)~dy~dx

    Now rewriting this as (this is so people who aren't super accustomed to double integrals can see what I am doing)

    \iint_{R}e^{-yx}\csc(px)~dA

    Where R=\left\{(x,y):0\leq{x}\leq\infty~{a\leq{y}\leq{b}  }\right\}

    From here it is obvious that this region is a rectangle, so Fubini's Theorem applies

    So

    \int_0^{\infty}\int_a^{b}e^{-yx}\csc(px)~dy~dx=\int_a^{b}\int_0^{\infty}e^{-yx}\csc(px)~dx~dy

    Now this inner integral is bothering me, I cannot seem to get it in the eight minutes I have tried

    parts-no deal

    Letting J(\theta)=\int_0^{\infty}e^{-yx}\csc(px)e^{-\theta\sin(px)}dx-no go

    Any suggestions?


    I also tried some crazy thing where I let

    J(\theta)=\int_0^{\infty}\frac{e^{-xy\cos(px\theta)}}{\sin(px)}~dx

    And found that the integral I want is a constant difference from J'(0)
    I believe that this is where you're screwing up...

    I think that it should be \int_0^{\infty}{\color{red}\sin(px)}\int_a^{b}e^{-yx}~dy~dx=\int_0^{\infty}\int_a^{b}e^{-yx}{\color{red}\sin(px)}~dy~dx

    because : \frac{e^{-ab}-e^{-bx}}{x\csc(px)}=\frac{e^{-ab}-e^{-bx}}{x}\cdot\frac{1}{\csc(px)}=\frac{(e^{-ab}-e^{-bx}){\color{red}\sin(px)}}{x}

    EDIT : icemanfan beat me to it...
    Follow Math Help Forum on Facebook and Google+

  4. #4
    MHF Contributor Mathstud28's Avatar
    Joined
    Mar 2008
    From
    Pennsylvania
    Posts
    3,641
    Quote Originally Posted by Chris L T521 View Post
    I believe that this is where you're screwing up...

    I think that it should be \int_0^{\infty}{\color{red}\sin(px)}\int_a^{b}e^{-yx}~dy~dx=\int_0^{\infty}\int_a^{b}e^{-yx}{\color{red}\sin(px)}~dy~dx

    because : \frac{e^{-ab}-e^{-bx}}{x\csc(px)}=\frac{e^{-ab}-e^{-bx}}{x}\cdot\frac{1}{\csc(px)}=\frac{(e^{-ab}-e^{-bx}){\color{red}\sin(px)}}{x}

    EDIT : icemanfan beat me to it...
    <-- with an f. Now I feel stupid for copying it down wrong

    Well

    \int{e^{-yx}{\color{red}\sin(px)}dx}=\left[\frac{-p\cos(px)}{y^2+p^2}-\frac{\sin(px)y}{y^2+p^2}\right]e^{-yx}

    By Integration by Parts twice then adding to both sides.

    So now as x goes to infinity obviously the e^{-yx} overpowers and it goes to zero, and when x=0 the above equals

    \frac{-p}{y^2+p^2}

    So we have that

    \int_0^{\infty}e^{-yx}\sin(px)~dx=\frac{p}{y^2+p^2}


    So now we must perform

    \int_a^{b}\frac{p}{x^2+p^2}~dx

    Rewriting this as

    \frac{1}{p}\int\frac{dx}{\left(\frac{x}{p}\right)^  2+1}~dx

    And letting u=\frac{x}{p}\Rightarrow{p\cdot{du}=dx}

    So we have

    \int\frac{du}{u^2+1}=\arctan(u)

    Back subbing we get

    \arctan\left(\frac{x}{p}\right)

    \therefore\quad\int_a^{b}\frac{p}{x^2+p^2}~dx=\arc  tan\left(\frac{b}{p}\right)-\arctan\left(\frac{a}{p}\right)


    \therefore\quad\boxed{\int_0^{\infty}\frac{e^{-ax}-e^{-bx}}{x\csc(px)}~dx=\arctan\left(\frac{b}{p}\right)-\arctan\left(\frac{a}{p}\right)}


    Seeing that I did a-b opposed to b-a this is what they had
    Follow Math Help Forum on Facebook and Google+

  5. #5
    Rhymes with Orange Chris L T521's Avatar
    Joined
    May 2008
    From
    Santa Cruz, CA
    Posts
    2,844
    Thanks
    3
    Quote Originally Posted by Mathstud28 View Post
    You know me well sir. I already was there deriving them, now most of them I can see what to do. But one is alluding me . Now I haven't given it much time but

    \int_0^{\infty}\frac{e^{-ax}-e^{-bx}}{x\csc(px)}

    Ok my first move was to do the obvious double integral

    \frac{e^{-ax}-e^{-bx}}{x}=\int_a^{b}e^{-yx}dy

    So we have

    \int_0^{\infty}\csc(px)\int_a^{b}e^{-yx}~dy~dx=\int_0^{\infty}\int_a^{b}e^{-yx}\csc(px)~dy~dx

    Now rewriting this as (this is so people who aren't super accustomed to double integrals can see what I am doing)

    \iint_{R}e^{-yx}\csc(px)~dA

    Where R=\left\{(x,y):0\leq{x}\leq\infty~{a\leq{y}\leq{b}  }\right\}

    From here it is obvious that this region is a rectangle, so Fubini's Theorem applies

    So

    \int_0^{\infty}\int_a^{b}e^{-yx}\csc(px)~dy~dx=\int_a^{b}\int_0^{\infty}e^{-yx}\csc(px)~dx~dy

    Now this inner integral is bothering me, I cannot seem to get it in the eight minutes I have tried

    parts-no deal

    Letting J(\theta)=\int_0^{\infty}e^{-yx}\csc(px)e^{-\theta\sin(px)}dx-no go

    Any suggestions?


    I also tried some crazy thing where I let

    J(\theta)=\int_0^{\infty}\frac{e^{-xy\cos(px\theta)}}{\sin(px)}~dx

    And found that the integral I want is a constant difference from J'(0)
    My 89 doesn't give a solution and Mathematica tells me this:



    Why don't you try to solve this using infinite series?

    \int_0^{\infty}\frac{e^{-ax}-e^{-bx}}{k\csc(px)}\,dx=\sum_{k=0}^{\infty}\left[e^{-ak}-e^{-bk}\right]\cdot\sum_{k=0}^{\infty}\left[\frac{\sin(k)}{k}\right]

    You know the power series expansions better than I do. Experiment with this and let me know how you do. I'll try to remember what I can to see I can evaluate the integral...

    Wait...can this be related to the zeta function, by any chance [I'm probably wrong...just throwing ideas out there]?

    --Chris
    Follow Math Help Forum on Facebook and Google+

  6. #6
    Rhymes with Orange Chris L T521's Avatar
    Joined
    May 2008
    From
    Santa Cruz, CA
    Posts
    2,844
    Thanks
    3
    Quote Originally Posted by Mathstud28 View Post
    <-- with an f. Now I feel stupid for copying it down wrong

    Well

    \int{e^{-yx}{\color{red}\sin(px)}dx}=\left[\frac{-p\cos(px)}{y^2+p^2}-\frac{\sin(px)y}{y^2+p^2}\right]e^{-yx}

    By Integration by Parts twice then adding to both sides.

    So now as x goes to infinity obviously the e^{-yx} overpowers and it goes to zero, and when x=0 the above equals

    \frac{-p}{y^2+p^2}

    So we have that

    \int_0^{\infty}e^{-yx}\sin(px)~dx=\frac{p}{y^2+p^2}


    So now we must perform

    \int_a^{b}\frac{p}{x^2+p^2}~dx

    Rewriting this as

    \frac{1}{p}\int\frac{dx}{\left(\frac{x}{p}\right)^  2+1}~dx

    And letting u=\frac{x}{p}\Rightarrow{p\cdot{du}=dx}

    So we have

    \int\frac{du}{u^2+1}=\arctan(u)

    Back subbing we get

    \arctan\left(\frac{x}{p}\right)

    \therefore\quad\int_a^{b}\frac{p}{x^2+p^2}~dx=\arc  tan\left(\frac{b}{p}\right)-\arctan\left(\frac{a}{p}\right)


    \therefore\quad\boxed{\int_0^{\infty}\frac{e^{-ax}-e^{-bx}}{x\csc(px)}~dx=\arctan\left(\frac{b}{p}\right)-\arctan\left(\frac{a}{p}\right)}


    Seeing that I did a-b opposed to b-a this is what they had
    Quote Originally Posted by Chris L T521 View Post
    My 89 doesn't give a solution and Mathematica tells me this:



    Why don't you try to solve this using infinite series?

    \int_0^{\infty}\frac{e^{-ax}-e^{-bx}}{k\csc(px)}\,dx=\sum_{k=0}^{\infty}\left[e^{-ak}-e^{-bk}\right]\cdot\sum_{k=0}^{\infty}\left[\frac{\sin(k)}{k}\right]

    You know the power series expansions better than I do. Experiment with this and let me know how you do. I'll try to remember what I can to see I can evaluate the integral...

    Wait...can this be related to the zeta function, by any chance [I'm probably wrong...just throwing ideas out there]?

    --Chris
    Um...ok...disregard my last post...

    These are interesting integrals!

    --Chris
    Follow Math Help Forum on Facebook and Google+

  7. #7
    MHF Contributor Mathstud28's Avatar
    Joined
    Mar 2008
    From
    Pennsylvania
    Posts
    3,641
    Quote Originally Posted by Chris L T521 View Post
    Um...ok...disregard my last post...

    These are interesting integrals!

    --Chris
    Don't feel bad, you may not have seen that, but I made up a new identity

    \frac{1}{\csc(x)}=\csc(x)
    Follow Math Help Forum on Facebook and Google+

  8. #8
    MHF Contributor Mathstud28's Avatar
    Joined
    Mar 2008
    From
    Pennsylvania
    Posts
    3,641
    The other one

    \boxed{1}\quad\int_0^{\infty}\frac{e^{-ax}-e^{-bx}}{x\sec(px)}~dx

    Ok once again seeing that

    \frac{e^{-ax}-e^{-bx}}{x}=\int_{a}^{b}e^{-yx}dy

    We have that \boxed{1} is equivalent to

    \boxed{2}\quad\int_0^{\infty}\int_a^{b}e^{-yx}\cos(px)~dy~dx

    Once again seeing the region of integration is a rectangle we may switch the integration order without changing the limits by Fubini's Theorem

    \boxed{2}=\int_a^{b}\int_0^{\infty}e^{-yx}\cos(px)~dx~dy

    Now once again by double integration by parts and algebra

    \int{e^{-yx}\cos(px)}dx=e^{-yx}\bigg[\frac{p\sin(px)}{y^2+p^2}-\frac{\cos(px)y}{y^2+p^2}\bigg]\quad\boxed{3}

    Now \lim_{x\to\infty}\boxed{3}=0

    and \boxed{3} evaluated at zero is \frac{-y}{y^2+p^2}


    so \boxed{4}\quad\int_0^{\infty}e^{-yx}\cos(px)~dx=\frac{y}{y^2+p^2}


    So using \boxed{4}

    We see that

    \int_a^b\int_0^{\infty}e^{-yx}\cos(px)~dx~dy=\int_a^{b}\frac{y}{y^2+p^2}~dy

    Now for \int\frac{y}{y^2+p^2}~dy

    If we let u=y^2+p then du=2y\Rightarrow\frac{du}{2}=y

    So we have

    \frac{1}{2}\int\frac{du}{u}=\frac{1}{2}\ln(u)

    So back subbing we get

    \frac{1}{2}\ln\left(y^2+p^2\right)

    So

    \int_a^{b}\frac{y}{y^2+p^2}=\frac{1}{2}\bigg[\ln(b^2+p^2)-\ln(a^2+p^2)\bigg]=\frac{1}{2}\ln\left(\frac{b^2+p^2}{a^2+p^2}\right  )=\ln\left(\sqrt{\frac{b^2+p^2}{a^2+p^2}}\right)\q  uad\boxed{5}


    Now putting \boxed{1},\boxed{2},\boxed{3},\boxed{4},\boxed{5} together we get

    \boxed{\int_0^{\infty}\frac{e^{-ax}-e^{-bx}}{x\sec(px)}~dx=\ln\left(\sqrt{\frac{b^2+p^2}{a  ^2+p^2}}\right)}\quad\blacksquare
    Follow Math Help Forum on Facebook and Google+

  9. #9
    Math Engineering Student
    Krizalid's Avatar
    Joined
    Mar 2007
    From
    Santiago, Chile
    Posts
    3,654
    Thanks
    12
    Mmm... but express the integrand into complex form, the rest follows. (This is worth than applyin' partial integration twice.)
    Follow Math Help Forum on Facebook and Google+

  10. #10
    MHF Contributor Mathstud28's Avatar
    Joined
    Mar 2008
    From
    Pennsylvania
    Posts
    3,641
    \int_0^{1}\frac{\ln(1+x)}{x}~dx\quad

    So seeing that since we can rewrite this as
    \int_0^{1}\frac{dx}{x}\cdot\sum_{n=0}^{\infty}\fra  c{(-1)^nx^{n+1}}{n+1}

    =\int_0^{1}\sum_{n=0}^{\infty}\frac{(-1)x^{n}}{n+1}

    =\sum_{n=0}^{\infty}\frac{(-1)^nx^{n+1}}{(n+1)^2}\bigg|_0^{1}

    =\sum_{n=0}^{\infty}\frac{(-1)^n}{(n+1)^2}

    =\sum_{n=1}^{\infty}\frac{(-1)^{n-1}}{n^2}

    Now we know that

    \left(1-2^{1-x}\right)\zeta(x)=\sum_{n=1}^{\infty}\frac{(-1)^{n-1}}{n^x}

    So we can see that

    \sum_{n=1}^{\infty}\frac{(-1)^{n-1}}{n^2}=\left(1-2^{1-2}\right)\zeta(2)

    =\frac{1}{2}\cdot\frac{\pi^2}{6}

    =\frac{\pi^2}{12}


    \therefore\quad\boxed{\int_0^{1}\frac{\ln(1+x)}{x}  ~dx=\frac{\pi^2}{12}}


    I feel as though I have done this before...
    Follow Math Help Forum on Facebook and Google+

  11. #11
    MHF Contributor Mathstud28's Avatar
    Joined
    Mar 2008
    From
    Pennsylvania
    Posts
    3,641
    I want to do this one integral, but first I need to do another one

    \int_0^{1}\frac{\ln(x)}{x+1}~dx

    =\int_0^{1}\ln(x)\sum_{n=0}^{\infty}(-1)^nx^n~dx

    =\sum_{n=0}^{\infty}(-1)^n\cdot\int_0^{1}x^{n}\ln(x)~dx

    =\sum_{n=0}^{\infty}(-1)^n\cdot\bigg[\left(\frac{x\ln(x)}{n+1}-\frac{x}{(n+1)^2}\right)x^{n}\bigg]\bigg|_0^{1}

    =\sum_{n=0}^{\infty}(-1)^n\cdot\frac{-1}{(n+1)^2}

    =-\sum_{n=1}^{\infty}\frac{(-1)^{n-1}}{n^2}

    =-\left(1-2^{1-2}\right)\cdot\frac{\pi^2}{6}=\frac{-\pi^2}{12}

    \therefore\quad\int_0^1\frac{\ln(x)}{x+1}~dx=\frac  {-\pi^2}{12}\quad\boxed{1}

    So now that that ancillary integral is out of the way I will move onto the integral of interest


    \int_0^{1}\ln(x)\ln(1+x)~dx

    First we see that


    \ln(1+x)=\int_0^x\frac{dy}{1+y}

    So we have that

    \int_0^{1}\ln(x)\ln(1+x)~dx=\int_0^{1}\int_0^{x}\f  rac{\ln(x)}{1+y}~dy~dx

    =\int_0^{1}\int_y^{1}\frac{\ln(x)}{y+1}~dx~dy

    =\int_0^{1}\bigg[\frac{x\ln(x)}{y+1}-\frac{x}{y+1}\bigg]\bigg|_y^{1}~dy

    =\int_0^{1}\bigg[\frac{\ln(y)}{y+1}-\ln(y)-\frac{2}{y+1}+1\bigg]~dy

    =\int_0^{1}\frac{\ln(y)}{y+1}~dy+\int_0^{1}\bigg[-\ln(y)-\frac{2}{y+1}+1\bigg]~dy

    =\int_0^{1}\frac{\ln(y)}{y+1}~dy+\bigg[-y\ln(y)+y-2\ln|y+1|+y\bigg]\bigg|_0^{1}

    =\int_0^{1}\frac{\ln(y)}{y+1}~dy+2-2\ln(2)

    Now using \boxed{1}

    We can finally conclude that

    \boxed{\int_0^{1}\ln(x)\ln(1+x)~dx=2-2\ln(2)-\frac{\pi^2}{12}}

    Follow Math Help Forum on Facebook and Google+

  12. #12
    Super Member PaulRS's Avatar
    Joined
    Oct 2007
    Posts
    571
    Quote Originally Posted by Mathstud28 View Post
    \boxed{\int_0^{1}\ln(x)\ln(1+x)~dx=2-2\ln(2)-\frac{\pi^2}{12}}
    It may also be proved as follows

    <br />
\sum\limits_{k = 1}^{n} {\tfrac{{\left( { - 1} \right)^{k + 1} }}<br />
{k} \cdot x^k }  + R_n \left( x \right) = \ln \left( {x + 1} \right)<br />
where <br />
\left( { - 1} \right)^n  \cdot \int_0^x {\tfrac{{t^n }}<br />
{{1 + t}}} dt = R_n \left( x \right)<br />
thus, if x>0, we have: <br />
\left| {R_n \left( x \right)} \right| \leqslant \tfrac{{x^n }}<br />
{n} = \int_0^x {\tfrac{{t^n }}<br />
{t}} dt<br />

    <br />
\sum\limits_{k = 1}^{n} {\tfrac{{\left( { - 1} \right)^{k + 1} }}<br />
{k} \cdot x^k \cdot{\ln(x)} }  +\ln(x)\cdot R_n \left( x \right) = \ln(x)\cdot \ln \left( {x + 1} \right)<br />
where <br />
\left| {\ln \left( x \right) \cdot R_n \left( x \right)} \right| \leqslant  - \tfrac{{x^n }}<br />
{n} \cdot \ln \left( x \right)<br />
if x\in(0, 1]

    Integrate, recalling the linearity of the integral and <br />
\int_0^1 {x^k  \cdot \ln \left( x \right)dx}  =  - \tfrac{1}<br />
{{\left( {k + 1} \right)^2 }}<br />

    <br />
\sum\limits_{k = 1}^n {\tfrac{{\left( { - 1} \right)^k }}<br />
{{k \cdot \left( {k + 1} \right)^2 }}}  + \int_0^1 {\ln \left( x \right) \cdot R_n \left( x \right)dx}  = \int_0^1 {\ln \left( x \right) \cdot \ln \left( {x + 1} \right)dx} <br />
(1)

    <br />
\left| {\int_0^1 {\ln \left( x \right) \cdot R_n \left( x \right)dx} } \right| \leqslant \int_0^1 {\left| {\ln \left( x \right) \cdot R_n \left( x \right)} \right|dx}  \leqslant  - \tfrac{1}<br />
{n}\int_0^1 {x^n  \cdot \ln \left( x \right)dx} <br />
thus: <br />
\left| {\int_0^1 {\ln \left( x \right) \cdot R_n \left( x \right)dx} } \right| \leqslant \tfrac{1}<br />
{{n \cdot \left( {n + 1} \right)^2 }}<br />

    Thus <br />
\left| {\int_0^1 {\ln \left( x \right) \cdot R_n \left( x \right)dx} } \right| \to 0<br />
as n tends to infinity and then <br />
\sum\limits_{k = 1}^\infty  {\tfrac{{\left( { - 1} \right)^k }}<br />
{{k \cdot \left( {k + 1} \right)^2 }}}  = \int_0^1 {\ln \left( x \right) \cdot \ln \left( {x + 1} \right)dx} <br />

    Now note that: <br />
\tfrac{{\left( { - 1} \right)^k }}<br />
{{k \cdot \left( {k + 1} \right)^2 }} = \tfrac{{\left( { - 1} \right)^k }}<br />
{k} + \tfrac{{\left( { - 1} \right)^{k + 1} }}<br />
{{k + 1}} - \tfrac{{\left( { - 1} \right)^k }}<br />
{{\left( {k + 1} \right)^2 }}<br />
thus: <br />
\sum\limits_{k \geqslant 1} {\tfrac{{\left( { - 1} \right)^k }}<br />
{{k \cdot \left( {k + 1} \right)^2 }}}  = \sum\limits_{k \geqslant 1} {\tfrac{{\left( { - 1} \right)^k }}<br />
{k}}  + \sum\limits_{k \geqslant 1} {\tfrac{{\left( { - 1} \right)^{k + 1} }}<br />
{{k + 1}}}  - \sum\limits_{k \geqslant 1} {\tfrac{{\left( { - 1} \right)^k }}<br />
{{\left( {k + 1} \right)^2 }}} <br />
(summed in that order according to (1))

    Now since : <br />
\sum\limits_{k \geqslant 1} {\tfrac{{\left( { - 1} \right)^k }}<br />
{k}}  =  - \ln \left( 2 \right);\sum\limits_{k \geqslant 1} {\tfrac{{\left( { - 1} \right)^{k + 1} }}<br />
{{k + 1}}}  =  - \ln \left( 2 \right) + 1<br />
and <br />
\sum\limits_{k \geqslant 1} {\tfrac{{\left( { - 1} \right)^k }}<br />
{{\left( {k + 1} \right)^2 }}}  = 1 - \tfrac{{\pi ^2 }}<br />
{{12}}<br />
we get the result
    Follow Math Help Forum on Facebook and Google+

  13. #13
    MHF Contributor Mathstud28's Avatar
    Joined
    Mar 2008
    From
    Pennsylvania
    Posts
    3,641
    Nice Paul!

    Could someone give me a starting hint on the two following ones?

    \int_0^{\infty}\frac{\sin^2(x)}{x^2}~dx

    Still Lebniz's rule?

    \int_0^{\infty}\frac{\cos(bx)-\cos(ax)}{x}~dx

    I would say double integration but I dont know the cosine of infinity
    Follow Math Help Forum on Facebook and Google+

  14. #14
    MHF Contributor Mathstud28's Avatar
    Joined
    Mar 2008
    From
    Pennsylvania
    Posts
    3,641
    Quote Originally Posted by Mathstud28 View Post
    \boxed{\int_0^{\infty}\frac{e^{-ax}-e^{-bx}}{x\sec(px)}~dx=\ln\left(\sqrt{\frac{b^2+p^2}{a  ^2+p^2}}\right)}\quad\blacksquare
    I thought this was kind of coincidental, this ones not on the list so I hope I am correct.

    \int_0^{\infty}\frac{\cos(ax)-\cos(bx)}{e^{px}x}~dx

    =\int_0^{\infty}e^{-px}\int_a^{b}\sin(yx)~dy~dx

    =\int_0^{\infty}\int_a^{b}e^{-px}\sin(yx)~dy~dx

    =\int_a^{b}\int_0^{\infty}e^{-px}\sin(yx)~dx~dy

    =\int_a^{b}\bigg[\frac{-e^{-px}\cos(xy)y}{y^2+p^2}-\frac{pe^{-px}\sin(xy)}{y^2+p^2}\bigg]\bigg|_0^{\infty}~dy

    =\int_a^{b}\frac{y}{y^2+p^2}~dy

    =\frac{1}{2}\bigg[\ln(b^2+p^2)-\ln(a^2+p^2)\bigg]

    =\boxed{\ln\left(\sqrt{\frac{b^2+p^2}{a^2+p^2}}\ri  ght)}

    \Rightarrow\int_0^{\infty}\cos(px)\frac{e^{-ax}-e^{-bx}}{x}~dx=\int_0^{\infty}e^{-px}\frac{\cos(ax)-\cos(bx)}{x}~dx


    That blows my mind, I hope I am right, otherwise this will be dissapointing.
    Follow Math Help Forum on Facebook and Google+

  15. #15
    Math Engineering Student
    Krizalid's Avatar
    Joined
    Mar 2007
    From
    Santiago, Chile
    Posts
    3,654
    Thanks
    12
    Quote Originally Posted by Mathstud28 View Post

    Could someone give me a starting hint on the two following ones?

    \int_0^{\infty}\frac{\sin^2(x)}{x^2}~dx
    Express \sin^2x into another form by using \cos2x. Consider that \frac{1}{x^{2}}=\int_{0}^{\infty }{ye^{-xy}\,dy}.

    Quote Originally Posted by Mathstud28 View Post

    \int_0^{\infty}\frac{\cos(bx)-\cos(ax)}{x}~dx

    I would say double integration but I dont know the cosine of infinity
    You recently found a parameter for the integrand.
    Follow Math Help Forum on Facebook and Google+

Page 1 of 6 12345 ... LastLast

Similar Math Help Forum Discussions

  1. Contour Integrals (to Evaluate Real Integrals)
    Posted in the Differential Geometry Forum
    Replies: 2
    Last Post: January 17th 2011, 09:23 PM
  2. Replies: 1
    Last Post: December 6th 2009, 07:43 PM
  3. Integrals : 2
    Posted in the Calculus Forum
    Replies: 4
    Last Post: November 24th 2009, 07:40 AM
  4. Integrals and Indefinite Integrals
    Posted in the Calculus Forum
    Replies: 3
    Last Post: November 9th 2009, 04:52 PM
  5. integrals Help please
    Posted in the Calculus Forum
    Replies: 2
    Last Post: May 8th 2008, 06:16 PM

Search Tags


/mathhelpforum @mathhelpforum