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Math Help - Integrals

  1. #46
    MHF Contributor Mathstud28's Avatar
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    I am actually going to use induction on this one, so if I mess it up please forgive me.

    \int_0^{1}\ln^n(x)~dx=(-1)^nn!\quad{n\in\mathbb{N}}

    Base Case

    Let us start with the base case n=0

    \int_0^1\ln^0(x)~dx

    =\int_0^1~dx

    =1

    =(-1)^00!

    So we have proved the base case.

    Inductive Hypothesis

    Now lets assume that

    \int_0^1\ln^n(x)~dx=(-1)^nn!\quad{n\in\mathbb{N}}

    This being our inductive hypothesis (IH)

    Inductive step

    Now lets look at our inductive step

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

    Which by Integration by Parts

    =x\ln^{n+1}(x)\bigg|_0^1-\int_0^{1}\frac{(n+1)\ln(x)^nx}{x}~dx

    =0-\int_0^1(n+1)\ln^n(x)~dx

    =-\bigg[n\int_0^1\ln^n(x)~dx+\int_0^1\ln(x)^n~dx\bigg]

    \underbrace{=}_{\text{IH}}=-\bigg[n(-1)^nn!+(-1)^nn!\bigg]

    =-(-1)^nn!(n+1)

    =(-1)^{n+1}(n+1)!

    Thus our inductive step is proven.

    \therefore\quad\boxed{\int_0^1\ln^n(x)~dx=(-1)^nn!\quad{n\in\mathbb{N}}}\quad\blacksquare



    We can make a stronger case of this

    \int_0^1\ln^n(x)~dx=(-1)^n\Gamma(n+1)\quad{n\in\mathbb{R}}


    This is gotten by simple aglaebraic manipulation

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

    \Rightarrow\int_0^1(-1)^n\ln^n(x)~dx=\Gamma(n+1)

    \Rightarrow{\int_0^1\left[\ln\left(\frac{1}{x}\right)\right]^n~dx=\Gamma(n+1)}


    Now on the left hand integral if we let -\ln(x)=t

    We get that

    x=e^{-t}

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

    and as x\to{0}\Rightarrow{t\to\infty}\

    And as x\to{1}\Rightarrow{t\to{0}}

    So we get

    -\int_{\infty}^{0}t^ne^{-t}~dt=\Gamma(n+1)

    =\int_0^{\infty}t^{n+1-1}e^{-t}~dt=\Gamma(n+1)

    \Gamma(n+1)=\Gamma(n+1)\quad\blacksquare
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  2. #47
    MHF Contributor Mathstud28's Avatar
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    Here is one I found implication but not by itself

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

    We first do this by seeing that

    \int\frac{dx}{1-x^2}

    =\int\sum_{n=0}^{\infty}x^{2n}~dx

    \sum_{n=0}^{\infty}\frac{x^{2n+1}}{2n+1}

    ={\rm{arctanh}}(x)\quad\forall{x}\backepsilon|x|<1

    so


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

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

    =\sum_{n=0}^{\infty}\frac{1}{2n+1}\int_0^1x^{2n}~d  x

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

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

    =\frac{\pi^2}{8}


    \therefore\quad\boxed{\int_0^1\frac{{\rm{arctanh}}  (x)}{x}~dx=\frac{\pi^2}{8}}\quad\blacksquare








    This brings a much simpler calculations of

    \int_0^1\frac{\ln(x)}{1-x^2}~dx

    Because

    \ln(x)=\int_1^x\frac{dy}{y}


    So we have


    \int_0^1\int_1^x\frac{dy}{y(1-x^2)}~dx

    =-\int_0^1\int_0^y\frac{dx}{y(1-x^2)}~dy

    =-\int_0^1\frac{dy}{y}\cdot\bigg[{\rm{arctanh}}(x)\bigg]\bigg|_0^1

    =-\int_0^1\frac{{\rm{arctanh}}(y)}{y}~dy

    =\frac{-\pi^2}{8}
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  3. #48
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    Quote Originally Posted by Mathstud28 View Post
    Here is one I found implication but not by itself

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

    We first do this by seeing that

    \int\frac{dx}{1-x^2}

    =\int\sum_{n=0}^{\infty}x^{2n}~dx

    \sum_{n=0}^{\infty}\frac{x^{2n+1}}{2n+1}

    ={\rm{arctanh}}(x)\quad\forall{x}\backepsilon|x|<1

    so


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

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

    =\sum_{n=0}^{\infty}\frac{1}{2n+1}\int_0^1x^{2n}~d  x

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

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

    =\frac{\pi^2}{8}


    \therefore\quad\boxed{\int_0^1\frac{{\rm{arctanh}}  (x)}{x}~dx=\frac{\pi^2}{8}}\quad\blacksquare








    This brings a much simpler calculations of

    \int_0^1\frac{\ln(x)}{1-x^2}~dx

    Because

    \ln(x)=\int_1^x\frac{dy}{y}


    So we have


    \int_0^1\int_1^x\frac{dy}{y(1-x^2)}~dx

    =-\int_0^1\int_0^y\frac{dx}{y(1-x^2)}~dy

    =-\int_0^1\frac{dy}{y}\cdot\bigg[{\rm{arctanh}}(x)\bigg]\bigg|_0^1

    =-\int_0^1\frac{{\rm{arctanh}}(y)}{y}~dy

    =\frac{-\pi^2}{8}
    Mathstud28, it a really good work! I find it fantastic how you evaluate integrals

    (just a small note: the inverse functions of the hyperbolic functions are the area-functions, not arcus functions (as you know of course), which the abbr. name is arsinh, arcosh, artanh, arcoth, arsech, arcsch of - all I mean is, it's not arc... but ar... )

    Inverse hyperbolic function - Wikipedia, the free encyclopedia

    I pretty much like the notaions arc and ar instead of that one with ^-1 cuz the 2nd one is ambiguous - you can think of it as 1/function

    best regards, marine
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  4. #49
    Super Member PaulRS's Avatar
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    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)
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  5. #50
    MHF Contributor Mathstud28's Avatar
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    \int_0^{\frac{\pi}{2}}\ln\left(\sin(x)\right)~dx

    This one is the one that has caused me the most mental anguish. I had nowhere to start and I just layed in bed and thought about it for twenty minutes and then went back to the drawing board. ad infinitum. It was easy once I realized something.

    If we make a substitution

    Let x=\frac{\pi}{2}-r\Rightarrow{t=\frac{\pi}{2}-x}

    \Rightarrow{dx=-dt}



    Then as x\to{0}\Rightarrow{t\to{\frac{\pi}{2}}}

    and as x\to\frac{\pi}{2}\Rightarrow{t\to{0}}

    So we get

    -\int_{\frac{\pi}{2}}^0\ln\left(\sin\left(\tiny{\fr  ac{\pi}{2}-t}\right)\right)~dt

    =\int_0^{\frac{\pi}{2}}\ln\left(\cos(t)\right)~dt


    \Rightarrow\quad\int_0^{\frac{\pi}{2}}\ln\left(\si  n(x)\right)~dx=\int_0^{\frac{\pi}{2}}\ln\left(\cos  (x)\right)~dx


    So now using this we see the following

    2\int_0^{\frac{\pi}{2}}\ln\left(\sin(x)\right)~dx

    =\int_0^{\frac{\pi}{2}}\ln\left(\sin(x)\right)~dx+  \int_0^{\frac{\pi}{2}}\ln\left(\cos(x)\right)~dx

    \int_0^{\frac{\pi}{2}}\ln\left(\cos(x)\sin(x)\righ  t)~dx

    Now I wanted \sin(2x) so I did the following

    =\int_0^{\frac{\pi}{2}}\ln\left(\sin(2x)\right)~dx-\int_0^{\frac{\pi}{2}}\ln(2)~dx

    Now for the first integral, if you let

    2x=t you get

    =\frac{1}{2}\int_0^{\pi}\ln\left(\sin(x)\right)~dx-\int_0^{\frac{\pi}{2}}\ln(2)~dx

    Now once again for this first integral let

    u=t-\frac{\pi}{2}

    \Rightarrow{dt=du}

    And as t\to{0}\Rightarrow{u\to\frac{-\pi}{2}}

    And as t\to{\pi}\Rightarrow{u\to\frac{\pi}{2}}

    Giving us

    =\frac{1}{2}\int_{\frac{-\pi}{2}}^{\frac{\pi}{2}}\ln\left(\sin\left(u+\frac  {\pi}{2}\right)\right)~du-\int_0^{\frac{\pi}{2}}\ln(2)~dx

    =\frac{1}{2}\int_{\frac{-\pi}{2}}^{\frac{\pi}{2}}\ln\left(\cos(u)\right)~du-\int_0^{\frac{\pi}{2}}\ln(2)~dx

    Now seeing that since

    \cos(-x)=\cos(x)

    \Rightarrow\ln(\cos(-x))=\ln(\cos(x))

    Which means that

    =\int_0^{\frac{\pi}{2}}\ln\left(\cos(u)\right)~du-\int_0^{\frac{\pi}{2}}\ln(2)~dx

    =\int_0^{\frac{\pi}{2}}\ln\left(\sin(x)\right)~dx-\int_0^{\frac{\pi}{2}}\ln(2)~dx

    So


    2\int_0^{\frac{\pi}{2}}\ln\left(\sin(x)\right)~dx=  \int_0^{\frac{\pi}{2}}\sin(x)~dx-\int_0^{\frac{\pi}{2}}\ln(2)~dx

    \Rightarrow\int_0^{\frac{\pi}{2}}\ln\left(\sin(x)\  right)~dx=-\int_0^{\frac{\pi}{2}}\ln(2)

    =\frac{-\ln(2)\pi}{2}


    So FINALLY we can conclude that


    \boxed{\int_0^{\frac{\pi}{2}}\ln\left(\sin(x)\righ  t)~dx=\int_0^{\frac{\pi}{2}}\ln\left(\cos(x)\right  )~dx=\frac{-\ln(2)\pi}{2}}\quad\blacksquare



    That one was a toughie.
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  6. #51
    MHF Contributor Mathstud28's Avatar
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    Quote Originally Posted by Marine View Post
    Mathstud28, it a really good work! I find it fantastic how you evaluate integrals

    (just a small note: the inverse functions of the hyperbolic functions are the area-functions, not arcus functions (as you know of course), which the abbr. name is arsinh, arcosh, artanh, arcoth, arsech, arcsch of - all I mean is, it's not arc... but ar... )

    Inverse hyperbolic function - Wikipedia, the free encyclopedia

    I pretty much like the notaions arc and ar instead of that one with ^-1 cuz the 2nd one is ambiguous - you can think of it as 1/function

    best regards, marine
    Thank you! I actually always thought it was arc . Well..I think I am going to keep using it ...why break convention for something as petty as looking stupid
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  7. #52
    MHF Contributor Mathstud28'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)
    I'll try .. \int_0^{\frac{\pi}{2}}\frac{\arctan\left(\sin(x)\r  ight)}{\sin(x)}~dx

    \Rightarrow\int_0^1\frac{\arctan(u)}{u\sqrt{1-u^2}}~du

    \sum_{n=0}^{\infty}\frac{(-1)^n}{2n+1}\int_0^1\frac{u^{2n}}{\sqrt{1-u^2}}

    =\sum_{n=0}^{\infty}\frac{\sqrt{\pi}(-1)^n\Gamma\left(\frac{2n+1}{2}\right)}{2\Gamma\lef  t(\frac{2n+1}{2}+\frac{3}{2}\right)}

    ..........................

    =\frac{\pi}{4}\ln\left(\frac{\sqrt{2}+1}{\sqrt{2}-1}\right)





    I will give it some more serious thought later.
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  8. #53
    Super Member PaulRS's Avatar
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    Quote Originally Posted by Mathstud28 View Post
    \int_0^{\frac{\pi}{2}}\ln\left(\sin(x)\right)~dx

    This one is the one that has caused me the most mental anguish. I had nowhere to start and I just layed in bed and thought about it for twenty minutes and then went back to the drawing board. ad infinitum. It was easy once I realized something.
    Here's another way I've found of computing it...

    Consider the Series: <br />
\sum\limits_{k = 1}^\infty  {\tfrac{{\cos \left( {k \cdot x} \right)}}<br />
{k}}  =  - \ln \left( 2 \right) - \ln \left[ {\sin \left( {\tfrac{x}<br />
{2}} \right)} \right]<br />
for all x such that <br />
0 < x < 2\pi <br />

    Integrate from 0 to \pi now: <br />
\sum\limits_{k = 1}^\infty  {\tfrac{{\int_0^\pi  {\cos \left( {k \cdot x} \right)dx} }}<br />
{k}}  =  - \ln \left( 2 \right) \cdot \pi  - \int_0^\pi  {\ln \left[ {\sin \left( {\tfrac{x}<br />
{2}} \right)} \right]dx} <br />
(assuming we can do that)

    But note that the series on the LHS is 0, because each of the terms is 0.

    THus <br />
0 =  - \ln \left( 2 \right) \cdot \pi  - \int_0^\pi  {\ln \left[ {\sin \left( {\tfrac{x}<br />
{2}} \right)} \right]dx} <br />

    From there it follows easily that: <br />
\int_0^{\frac{\pi}{2}}  {\ln \left[ {\sin \left( x \right)} \right]dx}  =  - \tfrac{\pi }<br />
{2} \cdot \ln \left( 2 \right)<br />
    Last edited by PaulRS; July 21st 2008 at 03:54 PM. Reason: a 1/2 was missing
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  9. #54
    MHF Contributor Mathstud28's Avatar
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    Quote Originally Posted by PaulRS View Post
    Here's another way I've found of computing it...

    Consider the Series: <br />
\sum\limits_{k = 1}^\infty {\tfrac{{\cos \left( {k \cdot x} \right)}}<br />
{k}} = - \ln \left( 2 \right) - \ln \left[ {\sin \left( {\tfrac{x}<br />
{2}} \right)} \right]<br />
for all x such that <br />
0 < x < 2\pi <br />

    Integrate from 0 to \pi now: <br />
\sum\limits_{k = 1}^\infty {\tfrac{{\int_0^\pi {\cos \left( {k \cdot x} \right)dx} }}<br />
{k}} = - \ln \left( 2 \right) \cdot \pi - \int_0^\pi {\ln \left[ {\sin \left( {\tfrac{x}<br />
{2}} \right)} \right]dx} <br />
(assuming we can do that)

    But note that the series on the LHS is 0, because each of the terms is 0.

    THus <br />
0 = - \ln \left( 2 \right) \cdot \pi - \int_0^\pi {\ln \left[ {\sin \left( {\tfrac{x}<br />
{2}} \right)} \right]dx} <br />

    From there it follows easily that: <br />
\int_0^\pi {\ln \left[ {\sin \left( x \right)} \right]dx} = - \tfrac{\pi }<br />
{2} \cdot \ln \left( 2 \right)<br />
    That is assuming the first identity is true (which I am sure it is)...but that is not a common identity...is it?
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  10. #55
    Moo
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    but that is not a common identity...is it?
    <br />
=\sum_{n=0}^{\infty}\frac{\sqrt{\pi}(-1)^n\Gamma\left(\frac{2n+1}{2}\right)}{2\Gamma\lef  t(\frac{2n+1}{2}+\frac{3}{2}\right)}<br />
    Ah..ah... :/

    And how do you get to this one ?

    By the way, \Gamma\left(\frac{2n+1}{2}+\frac{3}{2}\right)=\Gam  ma(n+2)=(n+1)!

    would it simplify ?
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  11. #56
    MHF Contributor Mathstud28's Avatar
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    Quote Originally Posted by Moo View Post
    Ah..ah... :/

    And how do you get to this one ?

    By the way, \Gamma\left(\frac{2n+1}{2}+\frac{3}{2}\right)=\Gam  ma(n+2)=(n+1)!

    would it simplify ?
    Yes sort of! Except the remaining Gamma function in the top .

    and as for your other question I don't know why but I remembered that

    \int_0^1\frac{x^{m}}{(1-x^n)^p}=\frac{\Gamma\left(\frac{m+1}{n}\right)\Gam  ma(p+1)}{\Gamma\left(\frac{m+1}{n}+p+1\right)}..B....e...t...
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  12. #57
    Moo
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    Quote Originally Posted by Mathstud28 View Post
    Yes sort of! Except the remaining Gamma function in the top .

    and as for your other question I don't know why but I remembered that

    \int_0^1\frac{x^{m}}{(1-x^n)^p}=\frac{\Gamma\left(\frac{m+1}{n}\right)\Gam  ma(p+1)}{\Gamma\left(\frac{m+1}{n}+p+1\right)}..B....e...t...
    erm...

    I know that the RHS is equal to \beta\left(\frac{m+1}{n}~,~p+1\right) (you cheater...you edited it lol !)
    And don't forget dx >.<

    You remembered that, but prove it, like for PaulRS !! lool
    According to my first thoughts on it, yours looks simpler than Paul's
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  13. #58
    MHF Contributor Mathstud28's Avatar
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    Quote Originally Posted by Moo View Post
    erm...

    (you cheater...you edited it lol !)
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  14. #59
    Eater of Worlds
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    I=\int_{0}^{\frac{\pi}{2}}ln(sin(x))dx

    Now, let's use the rather obscure identity:

    sin(x)=2sin(\frac{x}{2})cos(\frac{x}{2})

    I=\int_{0}^{\frac{\pi}{2}}ln\left(2sin(\frac{x}{2}  )cos(\frac{x}{2})\right)dx

    =\int_{0}^{\frac{\pi}{2}}ln(2)dx+\int_{0}^{\frac{\  pi}{2}}ln(sin(\frac{x}{2}))dx+\int_{0}^{\frac{\pi}  {2}}ln(cos(\frac{x}{2}))dx

    Make the sub u=\frac{x}{2}

    I=\frac{\pi}{2}ln(2)+2\int_{0}^{\frac{\pi}{4}}ln(s  in(u))du+2\int_{0}^{\frac{\pi}{4}}ln(cos(u))du......[1]

    Now, use the identity: cos(u)=sin(\frac{\pi}{2}-u)

    and make the sub t=\frac{\pi}{2}-u in the integral with the cos(u).

    We get:

    \int_{0}^{\frac{\pi}{4}}ln(cos(u))du

    = \int_{\frac{\pi}{4}}^{\frac{\pi}{2}}ln(sin(t))dt

    Then, [1] becomes:

    I=\frac{\pi}{2}ln(2)+2\int_{0}^{\frac{\pi}{4}}ln(s  in(u))du+2\int_{\frac{\pi}{4}}^{\frac{\pi}{2}}ln(s  in(t))dt

    I=\frac{\pi}{2}ln(2)+2\underbrace{\int_{0}^{\frac{  \pi}{2}}ln(sin(x))dx}_{\text{this is I}}

    I=\frac{\pi}{2}ln(2)+2I

    Solve for I:

    \boxed{I=\frac{-\pi}{2}ln(2)}
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  15. #60
    MHF Contributor Mathstud28's Avatar
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    \int_0^{\infty}\frac{\sin(ax)\cos(bx)}{x}~dx=I\qua  d{a\ne{b}}~~(a,b)>0

    We know that

    \sin(ax)\cos(bx)=\frac{1}{2}\bigg[\sin(ax-bx)+\sin(ax+bx)\bigg]

    and

    \frac{1}{x}=\int_0^{\infty}e^{-yx}~dy

    We see that

    I=\frac{1}{2}\int_0^{\infty}\bigg[\sin((a-b)x)e^{-yx}+e^{-yx}\sin((a+b)x)e^{-yx}\bigg]~dy~dx

    =\frac{1}{2}\int_0^{\infty}\int_0^{\infty}\bigg[\sin((a-b)x)e^{-yx}+\sin((a+b)x)e^{-yx}\bigg]~dx~dy

    Now let C=a-b and D=a+b

    Then we have

    \frac{1}{2}\int_0^{\infty}\int_0^{\infty}\bigg[\sin\left(Cx\right)e^{-yx}+\sin\left(Dx\right)e^{-yx}\bigg]~dx~dy

    Now when we evaluate this by parts I am sure since I have shown it many times that we get

    \frac{1}{2}\int_0^{\infty}\bigg[\frac{C}{y^2+C^2}+\frac{D^2}{y^2+D^2}\bigg]~dy

    =\frac{1}{2}\bigg[\arctan\left(\frac{y}{C}\right)+\arctan\left(\frac  {y}{D}\right)\bigg]\bigg|_0^{\infty}


    Now we want to back sub to get

    I=\frac{1}{2}\bigg[\arctan\left(\frac{y}{a-b}\right)+\arctan\left(\frac{y}{a+b}\right)\bigg]


    Now let us consider the two cases if

    a>b>0

    Then we have that

    I=\frac{1}{2}\bigg[\frac{\pi}{2}+\frac{\pi}{2}\bigg]

    =\frac{\pi}{2}


    Now if b>a>0

    This implies that a-b<0

    So by the oddness of arctangent

    I=\frac{1}{2}\bigg[\frac{-\pi}{2}+\frac{\pi}{2}\bigg]

    =0

    Now if a=b

    Then we go back to the beginning and rewite it as this

    \frac{1}{2}\int_0^{\infty}\frac{\sin(2ax)}{x}~dx

    Which we proved in another thread is equal to \frac{1}{2}\cdot\frac{\pi}{2}

    Giving us

    I=\frac{\pi}{4}

    Putting htis all together we get

    \boxed{\int_0^{\infty}\frac{\sin(ax)\cos(bx)}{x}~d  x=\begin{cases}<br />
0 & \mbox{if}~~0<a<b \\<br />
\frac{\pi}{2} & \mbox{if}~~0<b<a \\<br />
\frac{\pi}{4} &\mbox{if}~~a=b\ne{0}<br />
\end{cases} }
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