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  1. #61
    Super Member PaulRS's Avatar
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    Quote Originally Posted by PaulRS View Post
    Show that \int_0^{\frac{\pi}{2}}\frac{\arctan[\sin(x)]}{\sin(x)}dx=\frac{\pi}{4}\cdot{\ln\left(\frac{\sq  rt{2}+1}{\sqrt{2}-1}\right)}

    ( I don't have time to post the solution now. It's a nice integral, I like the result)
    OK, as promised.

    Lemma 1: <br />
\int_0^{\tfrac{\pi }<br />
{2}} {\sin ^{2n} \left( x \right)dx}  = \frac{\pi}{2^{2n+1}}\cdot{\binom{2n}{n}}<br />

    See here

    Lemma 2: <br />
\sum\limits_{n = 0}^\infty  {\binom{2n}{n} \cdot x^n }  = \tfrac{1}<br />
{{\sqrt {1 - 4x} }}<br />
whenever <br />
\left| x \right| < \tfrac{1}<br />
{4}<br />

    See here

    Computation: \int_0^{\frac{\pi}{2}}\frac{\arctan[\sin(x)]}{\sin(x)}dx=\frac{\pi}{4}\cdot{\ln\left(\frac{\sq  rt{2}+1}{\sqrt{2}-1}\right)}

    First we have: <br />
\tfrac{{\arctan \left[ {\sin \left( x \right)} \right]}}<br />
{{\sin \left( x \right)}} = 1 - \tfrac{{\sin ^2 \left( x \right)}}<br />
{3} + \tfrac{{\sin ^4 \left( x \right)}}<br />
{5} \mp ...<br />
for all <br />
\left| x \right| \leqslant \tfrac{\pi }<br />
{2}<br />
(just use the Taylor Series of \arctan(x) )

    Assuming uniform convergence : <br />
\int_0^{\tfrac{\pi }<br />
{2}} {\tfrac{{\arctan \left[ {\sin \left( x \right)} \right]}}<br />
{{\sin \left( x \right)}}dx}  = \int_0^{\tfrac{\pi }<br />
{2}} {dx}  - \tfrac{{\int_0^{\tfrac{\pi }<br />
{2}} {\sin ^2 \left( x \right)dx} }}<br />
{3} + \tfrac{{\int_0^{\tfrac{\pi }<br />
{2}} {\sin ^4 \left( x \right)dx} }}<br />
{5} \mp ...<br />

    By Lemma 1: <br />
\int_0^{\tfrac{\pi }<br />
{2}} {\tfrac{{\arctan \left[ {\sin \left( x \right)} \right]}}<br />
{{\sin \left( x \right)}}dx}  = \pi \cdot \sum\limits_{n = 0}^\infty  {\binom{2n}{n} \cdot \tfrac{{\left( { - 1} \right)^n }}<br />
{{2^{2n+1}  \cdot \left( {2n + 1} \right)}}} <br />

    By Lemma 2 : <br />
\sum\limits_{n = 0}^\infty  {\binom{2n}{n} \cdot \frac{(-1)^n\cdot x^{2n}}{2^{2n}} }  = \tfrac{1}<br />
{{\sqrt {1 + x^2} }}<br />

    Integrate from 0 to 1: 2\cdot<br />
\sum\limits_{n = 0}^\infty  {\binom{2n}{n} \cdot \tfrac{{\left( { - 1} \right)^n }}<br />
{{2^{2n+1}  \cdot \left( {2n + 1} \right)}}}  = \int_0^{1}\tfrac{dx}<br />
{{\sqrt {1 + x^2} }}=<br />
\left. {\text{arcsinh} \left( x \right)} \right|_0^1 =<br />
\tfrac{1}<br />
{2} \cdot \ln \left( {\tfrac{{\sqrt 2  + 1}}<br />
{{\sqrt 2  - 1}}} \right)<br /> <br />

    Which yields: \int_0^{\frac{\pi}{2}}\frac{\arctan[\sin(x)]}{\sin(x)}dx=\frac{\pi}{4}\cdot{\ln\left(\frac{\sq  rt{2}+1}{\sqrt{2}-1}\right)}
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  2. #62
    MHF Contributor Mathstud28's Avatar
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    \int\frac{e^{-ax}(1-\cos(px))}{x^2}~dx

    =\int_0^{\infty}\int_0^{\infty}ye^{-x(a+y)}(1-\cos(px))~dy~dx

    =\int_0^{\infty}\int_0^{\infty}ye^{-x(a+y)}(1-\cos(px))~dx~dy

    Now

    \int_0^{\infty}ye^{-x(y+a)}(1-\cos(px)~dx

    =y\bigg[\left(\frac{\cos(px)(y+a)}{(y+a)^2+p^2}-\frac{p\sin(px)}{(y+a)^2+p^2}+\frac{1}{y+a}\right)  e^{-x(y+a)}\bigg]\bigg|_0^{\infty}

    =\frac{-a}{y+a}+\frac{ay+a^2+p^2}{(y+a)^2+p^2}

    So we now need to evaluate

    \int_0^{\infty}\bigg[\frac{-a}{y+a}+\frac{ay+a^2+p^2}{(y+a)^2+p^2}\bigg]~dy

    =\int_0^{\infty}\bigg[\frac{-a}{y+a}+\frac{ay}{(y+a)^2+p^2}+\frac{a^2+p^2}{(y+a  )^2+p^2}\bigg]~dy


    The first term is obviously a\ln|a+y|

    Now let us examine the other terms

    \int\frac{ay}{(y+a)^2+p^2}~dy

    This can easily be gotten by letting

    z=y+a

    \Rightarrow{dz=dy}

    So we have

    \int\frac{a(z-a)}{z^2+p^2}

    =\int\bigg[a\frac{z}{z^2+p^2}-a^2\int\frac{1}{z^2+p^2}\bigg]~dz

    =\frac{a}{2}\ln(z^2+p^2)-\frac{a^2}{p}\arctan\left(\frac{z}{p}\right)+C

    \underbrace{=}_{\text{backsub}}\frac{a}{2}\ln\left  ((y+a)^2+p^2\right)-\frac{a^2}{p}\arctan\left(\frac{y+a}{p}\right)+C

    And

    \int\frac{a^2+p^2}{(y+a)^2+p^2}~dy

    Similarly with a substitution of z=y+a we get

    (a^2+p^2)\int\frac{dz}{z^2+p^2}

    =\frac{a^2+p^2}{p}\arctan\left(\frac{z}{p}\right)+  C

    \underbrace{=}_{\text{backsub}}=\frac{a^2+p^2}{p}\  arctan\left(\frac{y+a}{p}\right)+C

    So now putting this all back together we get

    \int\bigg[\frac{-a}{y+a}+\frac{ay}{(y+a)^2+p^2}-\frac{a^2+p^2}{(y+a)^2+p^2}\bigg]~dy

    =-a\ln|a+y|+\frac{a}{2}\ln\left((y+a)^2+p^2\right)-\frac{a^2}{p}\arctan\left(\frac{y+a}{p}\right) +\frac{a^2+p^2}{p}\arctan\left(\frac{y+a}{p}\right  )+C

    =-a\ln|a+y|+\frac{a}{2}\ln\left((y+a)^2+p^2\right)+p  \arctan\left(\frac{y+a}{p}\right)+C

    =F(y)

    So seeing that

    \int_0^{\infty}\bigg[\frac{a}{y+a}-\frac{ay+a^2+p^2}{(y+a)^2+p^2}\bigg]~dy=F\left(\infty\right)-F(0)

    We need to find

    F(\infty) and F(0)

    So we see that

    F(0)=\frac{a}{2}\ln\left(a^2+p^2\right)+p\arctan\l  eft(\frac{a}{p}\right)-a\ln|a|

    and

    F(\infty)=\frac{p{\pi}}{2}

    We put this together to get

    F\left(\infty\right)-F(0)=\frac{-a}{2}\ln(a^2+p^2)-p\arctan\left(\frac{a}{p}\right)+a\ln|a|+\frac{p\p  i}{2}

    =p\,{\rm{arccot}}\left(\frac{a}{p}\right)+a\ln\lef  t(\frac{|a|}{\sqrt{a^2+p^2}}\right)

    We can conclude that


    \boxed{\int_0^{\infty}\frac{e^{-ax}(1-\cos(px))}{x^2}~dx=p\,{\rm{arccot}}\left(\frac{a}{  p}\right)+a\ln\left(\frac{|a|}{\sqrt{a^2+p^2}}\rig  ht)}\quad\blacksquare
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  3. #63
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    Quote Originally Posted by PaulRS View Post
    OK, as promised.
    1)That is a really nice integral.

    2)Did you start using "\tfrac" because of me, it appears you are more obsessed with making you fractions smaller.

    3)How do you prove uniform convergence? Weierstrass test would not work here.
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  4. #64
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    Hi! Here are 2 improper integrals for you!

    1:

    Int[ Abs[ln[Abs[x]]]*a/(Abs[x])^3 * Exp(-a/x^2) ] dx from -oo to +oo

    2:

    Int[ a(Abs[x])^(i-3) * Exp[-a/x^2) ] dx from -oo to +oo

    i<2 ; a>0 for both integrals, and 'i' is NOT the imaginary unit

    Abs[x] - is the absolute value of x

    I would be thankful, if s.o. could type both integrals with TeX, so that I can check if everything's like it should be
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  5. #65
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    Hi! Here are 2 improper integrals for you!
    I will attempt to LaTex your integrals. I hope I am interpreting correctly.


    Int[ Abs[ln[Abs[x]]]*a/(Abs[x])^3 * Exp(-a/x^2) ] dx from -oo to +oo
    \int_{-\infty}^{\infty}\frac{a|ln|x||e^{\frac{-a}{x^{2}}}}{(|x|)^{3}}


    Int[ a(Abs[x])^(i-3) * Exp[-a/x^2) ] dx from -oo to +oo, i<2 ; a>0 for both integrals, and 'i' is NOT the imaginary unit
    I will use k instead of i.

    \int_{-\infty}^{\infty}a(|x|)^{k-3}\cdot e^{\frac{-a}{x^{2}}}, \;\ k<2, \;\ a>0
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  6. #66
    Moo
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    The function under the integral sign is even. Then we can write :



    Since x is positive, we can write :
    I=2 \int_0^{\infty} \frac{a |\ln(x)| e^{\frac{-a}{x^2}}}{x^3} \ dx


    I=2 \int_0^1 \frac{a |\ln(x)| e^{\frac{-a}{x^2}}}{x^3} \ dx+2 \int_1^{\infty} \frac{a |\ln(x)| e^{\frac{-a}{x^2}}}{x^3} \ dx


    Since ln(x)<0 for x between 0 & 1 and >0 for x>1 :

    I=-2 \int_0^1 \frac{a \ln(x) e^{\frac{-a}{x^2}}}{x^3} \ dx+2 \int_1^{\infty} \frac{a \ln(x) e^{\frac{-a}{x^2}}}{x^3} \ dx


    Substitute u=\frac{1}{x^2} \implies dx=-\frac 12 x^3 \ du

    x=sqrt(u) --> ln(x)=ln(u)/2

    The integral is now (after some simplifcations) :
    I=- \int_1^{\infty} a \cdot \frac{\ln(u)}{2} \cdot e^{-au} \ du+ \int_0^1 a \cdot \frac{\ln(u)}{2} \cdot e^{-au} \ du


    And now, I think an integration by parts can be done, but there's something worrying me
    Last edited by Moo; July 28th 2008 at 10:51 PM.
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  7. #67
    MHF Contributor Mathstud28's Avatar
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    Quote Originally Posted by Moo View Post
    The function under the integral sign is even. Then we can write :



    Since x is positive, we can write :
    I=2 \int_0^{\infty} \frac{a |\ln(x)| e^{\frac{-a}{x^2}}}{x^3} \ dx


    I=2 \int_0^1 \frac{a |\ln(x)| e^{\frac{-a}{x^2}}}{x^3} \ dx+2 \int_1^{\infty} \frac{a |\ln(x)| e^{\frac{-a}{x^2}}}{x^3} \ dx


    Since ln(x)<0 for x between 0 & 1 and >0 for x>1 :

    I=-2 \int_0^1 \frac{a \ln(x) e^{\frac{-a}{x^2}}}{x^3} \ dx+2 \int_1^{\infty} \frac{a \ln(x) e^{\frac{-a}{x^2}}}{x^3} \ dx


    Substitute u=\frac{1}{x^2} \implies dx=-\frac 12 x^3 \ du

    x=sqrt(u) --> ln(x)=ln(u)/2

    The integral is now (after some simplifcations) :
    I=- \int_1^{\infty} a \cdot \frac{\ln(u)}{2} \cdot e^{-au} \ du+ \int_0^1 a \cdot \frac{\ln(u)}{2} \cdot e^{-au} \ du


    And now, I think an integration by parts can be done, but there's something worrying me
    I haven't checked it, but assuming your work up to the last step is corect you cannot do integration by parts because natural log is not differentiably redundant. But what you can do is convert e^{-ax} to power series. Assume uniform convergence and switch integral and summation, do integratioy by parts on [math ]\int_0^1x^n\ln(x)~dx[/tex] and then go from there.

    You got it?
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  8. #68
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    I see you're having lot's of fun with them

    To be honest, I have an antiderivative for the first integral, nut this for the second one has series in it


    have fun, Marine
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  9. #69
    MHF Contributor Mathstud28's Avatar
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    Quote Originally Posted by Marine View Post
    I see you're having lot's of fun with them

    To be honest, I have an antiderivative for the first integral, nut this for the second one has series in it


    have fun, Marine
    THe one Moo did (the first one?) does not have an antiderivative composed entirely of elementary functions. Also the answer is -pi/4
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  10. #70
    MHF Contributor Mathstud28's Avatar
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    Quote Originally Posted by Moo View Post
    The function under the integral sign is even. Then we can write :



    Since x is positive, we can write :
    I=2 \int_0^{\infty} \frac{a |\ln(x)| e^{\frac{-a}{x^2}}}{x^3} \ dx


    I=2 \int_0^1 \frac{a |\ln(x)| e^{\frac{-a}{x^2}}}{x^3} \ dx+2 \int_1^{\infty} \frac{a |\ln(x)| e^{\frac{-a}{x^2}}}{x^3} \ dx


    Since ln(x)<0 for x between 0 & 1 and >0 for x>1 :

    I=-2 \int_0^1 \frac{a \ln(x) e^{\frac{-a}{x^2}}}{x^3} \ dx+2 \int_1^{\infty} \frac{a \ln(x) e^{\frac{-a}{x^2}}}{x^3} \ dx


    Substitute u=\frac{1}{x^2} \implies dx=-\frac 12 x^3 \ du

    x=sqrt(u) --> ln(x)=ln(u)/2

    The integral is now (after some simplifcations) :
    I=- \int_1^{\infty} a \cdot \frac{\ln(u)}{2} \cdot e^{-au} \ du+ \int_0^1 a \cdot \frac{\ln(u)}{2} \cdot e^{-au} \ du


    And now, I think an integration by parts can be done, but there's something worrying me
    Lets look at the first integral. If we let

    then we see that So we get








    Now assuming uniform convergence





    Now for the second integral once again letting



    We get







    So putting it all together we get



    I don't care enough to simplify, its too late. I am not sure that is right, its hard to switch back and forth from that Moo website to MHF and not make clarical errors. And this is all assuming Moo's integral was teh correct one.

    By the way in the step before the solution I didn't type a negative in front of the serise and there was no way i was going back to change it, its fixed in the final solution. And gamma should be negative
    Last edited by Mathstud28; July 23rd 2008 at 10:43 PM.
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  11. #71
    Moo
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    \int_0^1 \frac{\ln(1+x)}{1+x^2} \ dx

    Not seen in this thread, maybe in another one...?

    Integrals-8.gif
    Integrals-9.gif
    Integrals-10.gif
    Integrals-11.gif

    *code*


    Substitute ~ x=\tan \varphi \\
    I=\int_0^{\frac \pi 4} \ln \left(1+\tan \varphi\right) ~d \varphi \\
    I=\int_0^{\frac \pi 4} \ln \left(\cos \varphi+\sin \varphi\right)- \ln(\cos \varphi) ~d \varphi \\ \\

    Let~J=\int_0^{\frac \pi 4} \ln(\cos \varphi+\sin \varphi)~d\varphi \\
    We~can~observe~that~ \sin \left(x+\frac \pi 4 \right)=\frac{1}{\sqrt{2}}(\cos(x)+\sin(x)) \\
    Thus~\cos \varphi+\sin \varphi=\sqrt{2} \sin \left(\varphi+\frac \pi 4\right) \\ \\

    J=\int_0^{\frac \pi 4} \ln(\sqrt{2})+\ln \left(\sin \left(\varphi+\frac \pi 4\right)\right)~d\varphi \\
    J=\ln(2) \frac{ \pi }{ 8 } + \int_0^{\frac \pi 4} \ln \left(\sin \left(\varphi+\frac \pi 4\right)\right)~d\varphi \\
    Let~\phi=\frac \pi 4 - \varphi \Rightarrow \varphi+\frac \pi 4=\frac \pi 2-\phi \\
    J=\ln(2) \frac \pi 8 -\int_{\frac \pi 4}^0 \ln \left(\sin \left(\frac \pi 2-\phi\right)\right)~d \phi \\ \\
    J=\ln(2) \frac \pi 8+\int_0^{\frac \pi 4} \ln(\cos \phi) ~d\phi \\ \\ \\

    Therefore \ I=J-\int_0^{\frac \pi 4} \ln(\cos \varphi) \ d \varphi \\
    I=\ln(2) \frac \pi 8+\underbrace{\int_0^{\frac \pi 4} \ln(\cos \phi) \ d\phi-\int_0^{\frac \pi 4} \ln(\cos \varphi) \ d\varphi}_{=0} \\ \\
    I=\ln(2){\frac \pi 8}
    Last edited by Moo; July 28th 2008 at 10:50 PM.
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  12. #72
    MHF Contributor Mathstud28's Avatar
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    Quote Originally Posted by Moo View Post


    Not seen in this thread, maybe in another one...?

    *picture of the answer* (if you want to do it without seeing it)
    http://www.gnux.be/latex/data/972dbd...70b52fa1ad.png

    *code*


    Substitute ~ x=\tan \varphi \\
    I=\int_0^{\frac \pi 4} \ln \left(1+\tan \varphi\right) ~d \varphi \\
    I=\int_0^{\frac \pi 4} \ln \left(\cos \varphi+\sin \varphi\right)- \ln(\cos \varphi) ~d \varphi \\ \\
    Let~J=\int_0^{\frac \pi 4} \ln(\cos \varphi+\sin \varphi)~d\varphi \\
    We~can~observe~that~ \sin \left(x+\frac \pi 4 \right)=\frac{1}{\sqrt{2}}(\cos(x)+\sin(x)) \\
    Thus~\cos \varphi+\sin \varphi=\sqrt{2} \sin \left(\varphi+\frac \pi 4\right) \\ \\
    J=\int_0^{\frac \pi 4} \ln(\sqrt{2})+\ln \left(\sin \left(\varphi+\frac \pi 4\right)\right)~d\varphi \\
    J=\ln(2) \frac{ \pi }{ 8 } + \int_0^{\frac \pi 4} \ln \left(\sin \left(\varphi+\frac \pi 4\right)\right)~d\varphi \\
    Let~\phi=\frac \pi 4 - \varphi \Rightarrow \varphi+\frac \pi 4=\frac \pi 2-\phi \\
    J=\ln(2) \frac \pi 8 -\int_{\frac \pi 4}^0 \ln \left(\sin \left(\frac \pi 2-\phi\right)\right)~d \phi \\ \\
    J=\ln(2) \frac \pi 8+\int_0^{\frac \pi 4} \ln(\cos \phi) ~d\phi \\ \\ \\
    Therefore I=J-\int_0^{\frac \pi 4} \ln(\cos \varphi) ~d \varphi \\
    I=\ln(2) \frac \pi 8+\underbrace{\int_0^{\frac \pi 4} \ln(\cos \phi) ~d\phi-\int_0^{\frac \pi 4} \ln(\cos \varphi) ~d\varphi}_{=0} \\ \\
    I=\ln(2){\frac \pi 8}
    This is the integral I solved with the "coo" identity.

    My soultion is very simiar to yours except my trig identity was slightly different
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  13. #73
    MHF Contributor Mathstud28's Avatar
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    Ok I have a lot to do...lets start off nice and easy

    Number one

    \int_0^{\frac{\pi}{2}}\frac{dx}{1+\tan^{\alpha}(x)  }

    This one stumped me for a while unti you see something obvious

    Let

    I=\int_0^{\frac{\pi}{2}}\frac{dx}{1+\tan^{\alpha}(  x)}

    \underbrace{=}_{x\mapsto{\frac{\pi}{2}-z}}-\int_{\frac{\pi}{2}}^0\frac{dz}{1+\left(\frac{1}{\  tan(z)}\right)^{\alpha}}

    =\int_0^{\frac{\pi}{2}}\frac{\tan^{\alpha}(z)}{\ta  n^{\alpha}(z)+1}


    =I

    \Rightarrow{2}I=\int_0^{\frac{\pi}{2}}\frac{\tan^{  \alpha}(x)}{1+\tan^{\alpha}(x)}dx+\int_0^{\frac{\p  i}{2}}\frac{dx}{1+\tan^{\alpha}(x)}

    =\int_0^{\frac{\pi}{2}}dx

    =\frac{\pi}{2}

    2I=\frac{\pi}{2}

    \Rightarrow{I=\frac{\pi}{4}}

    \therefore \boxed{\int_0^{\frac{\pi}{2}}\frac{dx}{1+\tan^{\al  pha}(x)}=\frac{\pi}{4}}\quad\blacksquare
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  14. #74
    MHF Contributor Mathstud28's Avatar
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    This one is just plain easy but requires you to take something as true

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

    Now we know that

    \forall{x}\in[0,1]\arctan(x)\to\sum_{n=0}^{\infty}\frac{(-1)^nx^{2n+1}}{2n+1}~~\text{uniformly}

    \therefore\int_0^1\frac{\arctan(x)}{x}~dx

    =\int_0^1\frac{1}{x}\sum_{n=0}^{\infty}\frac{(-1)^nx^{2n+1}}{2n+1}~dx

    Which by virtue of uniform convergence

    =\sum_{n=0}^{\infty}\frac{(-1)^n}{2n+1}\int_0^1x^{2n}~dx

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

    =K

    Where K is Chatalan's constant \approx{.9159}

    \therefore\quad\boxed{\int_0^1\frac{\arctan(x)}{x}  ~dx=K}\quad\blacksquare
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  15. #75
    MHF Contributor Mathstud28's Avatar
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    I must prove this so I can use it later

    \int_0^1{x^m\ln^p(x)}~dx=\frac{(-1)^p\Gamma(p+1)}{(m+1)^{p+1}}

    This is very simple first let us consider a different integral

    \int_0^1{x^m\ln\left(\frac{1}{x}\right)^p~dx}

    Now let

    \ln\left(\frac{1}{x}\right)=z

    then

    x=e^{-z}

    \Rightarrow{dx=-e^{-z}}

    Now we see that the limits are changed

    0\mapsto\infty

    1\mapsto{0}


    So we get

    -\int_{\infty}^0e^{-mz}z^{p}\frac{dz}{e^{z}}

    =\int_0^{\infty}{e^{-z(m+1)}z^p}

    Now let t=z(m+1)

    \Rightarrow{\frac{t}{m+1}=z}

    \Rightarrow{\frac{dt}{m+1}=dz}

    Limits stay the same. So we get

    =\int_0^{\infty}e^{-t}\left(\frac{t}{m+1}\right)^p\frac{dt}{m+1}

    =\frac{1}{(m+1)^{p+1}}\int_0^{\infty}e^{-t}t^{(p+1)-1}~dt

    =\frac{\Gamma(p+1)}{(m+1)^{p+1}}

    \Rightarrow\int_0^1{x^m\ln\left(\frac{1}{x}\right)  ^p~dx}=\frac{\Gamma(p+1)}{(m+1)^{p+1}}

    Now notice that

    \int_0^{1}x^m\ln\left(\frac{1}{x}\right)^p~dx

    =\int_0^{1}x^m(-1)^p\ln^p(x)~dx

    So now dividing both sides appropriately gives us the final result

    \boxed{\int_0^1{x^m\ln^p(x)}~dx=\frac{(-1)^p\Gamma(p+1)}{(m+1)^{p+1}}}\quad\blacksquare
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    Replies: 2
    Last Post: May 8th 2008, 06:16 PM

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