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Thread: MHF Integral Bee!

  1. #46
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    Quote Originally Posted by simplependulum View Post
    Problem :

    $\displaystyle L^{-1} [ \sqrt{s + \sqrt{s^2 + a^2}} ] $

    Is it related to INTEGRAL ? If you think it is not , i still have another problem
    The interesting thing of the problem is

    $\displaystyle \sqrt{s + \sqrt{s^2 + a^2}} = \frac{1}{2} \sqrt{ 2s + 2\sqrt{(s-ai)(s+ai)}}$

    $\displaystyle = \frac{1}{2} \sqrt{ s + ai + 2\sqrt{s+ai}\sqrt{s-ai} + s - ai}$

    $\displaystyle = \frac{1}{2} ( \sqrt{ s+ai} + \sqrt{s-ai}) $
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  2. #47
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    Quote Originally Posted by Amer View Post
    What is the new integral ?? luobo you should post an integral .
    The problem is copied from the internet.

    The harmonic series
    $\displaystyle
    1-\frac{1}{2}+\frac{1}{3}-\frac{1}{4}+\frac{1}{5}-\frac{1}{6}+ ...
    $
    converges. Show that the rearranged series
    $\displaystyle
    1+\frac{1}{3}-\frac{1}{2}+\frac{1}{5}+\frac{1}{7}-\frac{1}{4}+\frac{1}{9}+\frac{1}{11}-\frac{1}{6}+...
    $
    also converges and find its limit
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  3. #48
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    Quote Originally Posted by luobo View Post
    The problem is copied from the internet.

    The harmonic series
    $\displaystyle
    1-\frac{1}{2}+\frac{1}{3}-\frac{1}{4}+\frac{1}{5}-\frac{1}{6}+ ...
    $
    converges. Show that the rearranged series
    $\displaystyle
    1+\frac{1}{3}-\frac{1}{2}+\frac{1}{5}+\frac{1}{7}-\frac{1}{4}+\frac{1}{9}+\frac{1}{11}-\frac{1}{6}+...
    $
    also converges and find its limit
    well, i don't know, this might be related to "integration" somehow but this thread is supposed to be about "techniques of integration"! so let's just keep it that way please.

    i don't want this thread to go in a different direction. cheers!

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

    I'm going to bend the rules a little bit and give you a few integrals instead of one. Pick whichever you like and share your solution with us if you feel like it:

    1) $\displaystyle \int_0^1 (1-x)e^{-x} \ln x \ dx$

    2) $\displaystyle \int_0^1 \frac{x(x+1)\sin (\ln x)}{\ln x} \ dx$

    3) $\displaystyle \int_0^{\infty}\cot^{-1}(ax) \cot^{-1}(bx) \ dx, \ \ a > 0, \ b > 0$

    4) $\displaystyle \int_0^{\pi} (\sin x)^a \sin(ax) \ dx, \ \ a > -1$

    5) $\displaystyle \int_0^{\infty} \sin (x^a) \ dx, \ \ a > 1.$ (the answer to this one is in terms of Gamma function!)

    Have fune!
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  4. #49
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    Quote Originally Posted by NonCommAlg View Post

    1) $\displaystyle \int_0^1 (1-x)e^{-x} \ln x \ dx$
    Rewrite the integral as $\displaystyle -\int_0^1\int_x^1\frac{(1-x)e^{-x}}t\,dt\,dx,$ after reverse integration order and by straightforward computations we'll get that the integral is $\displaystyle -\int_0^1e^{-t}\,dt=\frac1e-1.$
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  5. #50
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    Quote Originally Posted by NonCommAlg View Post

    2) $\displaystyle \int_0^1 \frac{x(x+1)\sin (\ln x)}{\ln x} \ dx$
    Start with this simple fact: $\displaystyle \frac{\sin(\ln x)}{\ln x}=\int_0^1\cos(t\ln x)\,dt.$ Thus the integral becomes (after reversing integration order) $\displaystyle \int_0^1\int_0^1x(x+1)\cos(t\ln x)\,dx\,dt,$ so it remains to compute $\displaystyle \int_0^1\left(\frac3{t^2+9}+\frac2{t^2+4}\right)\, dt$ and its value is $\displaystyle \arctan\left(\frac13\right)+\arctan\left(\frac12\r ight)$ which is the answer of the integral.

    ----------

    See below for full answer. I did this when I was in college, so no time to do anything else!
    Last edited by Krizalid; Aug 21st 2009 at 03:59 PM.
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  6. #51
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    Quote Originally Posted by Krizalid View Post
    Start with this simple fact: $\displaystyle \frac{\sin(\ln x)}{\ln x}=\int_0^1\cos(t\ln x)\,dt.$ Thus the integral becomes (after reversing integration order) $\displaystyle \int_0^1\int_0^1x(x+1)\cos(t\ln x)\,dx\,dt,$ so it remains to compute $\displaystyle \int_0^1\left(\frac3{t^2+9}+\frac2{t^2+4}\right)\, dt$ and its value is $\displaystyle \arctan\left(\frac13\right)+\arctan\left(\frac12\r ight)$ which is the answer of the integral.
    and: $\displaystyle \tan^{-1}\left(\frac13\right)+\tan^{-1}\left(\frac12\right)=\tan^{-1} \left(\frac{\frac{1}{2}+\frac{1}{3}}{1 - \frac{1}{6}} \right) = \frac{\pi}{4}.$


    ok, two down, three to go:


    3) $\displaystyle \int_0^{\infty}\cot^{-1}(ax) \cot^{-1}(bx) \ dx, \ \ a > 0, \ b > 0$

    4) $\displaystyle \int_0^{\pi} (\sin x)^a \sin(ax) \ dx, \ \ a > -1$

    5) $\displaystyle \int_0^{\infty} \sin (x^a) \ dx, \ \ a > 1.$
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  7. #52
    MHF Contributor Bruno J.'s Avatar
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    I attack $\displaystyle
    I=\int_0^{\infty} \sin (x^a) \ dx, \ \ a > 1.
    $

    First we substitute $\displaystyle x^a=t$. Then $\displaystyle ax^{a-1}dx=dt$, i.e. $\displaystyle dx = \frac{dt}{ax^{a-1}}=\frac{1}{a}{t^\frac{1-a}{a}}\: dt$.

    Now $\displaystyle \int_0^{\infty} \sin (x^a) \ dx = \frac{1}{a}\int_0^{\infty} \sin (t){t^\frac{1-a}{a}}\: dt = \frac{1}{2ia}\int_0^{\infty} (e^{it}-e^{-it}){t^\frac{1-a}{a}}\: dt$.

    Now consider $\displaystyle I_1=\frac{1}{2ia}\int_0^{\infty} e^{it}t^\frac{1-a}{a}\: dt$. Set $\displaystyle it=-u$, so $\displaystyle dt = i\: du$. Then the integral becomes

    $\displaystyle \frac{1}{2ia}\int_0^{\infty} e^{-u}(iu)^\frac{1-a}{a}\: i\: du = \frac{i^{\frac{1-a}{a}}}{2a}\int_0^{\infty} e^{-u}u^{\frac{1}{a}-1}\: du = \frac{i^{\frac{1-a}{a}}}{2a}\Gamma\left(\frac{1}{a}\right)$.

    Similarily, consider $\displaystyle I_2=\frac{-1}{2ia}\int_0^{\infty} e^{-it}t^\frac{1-a}{a}\: dt$. Set $\displaystyle it=u$, so $\displaystyle dt = -i\: du$. Then the integral becomes

    $\displaystyle \frac{-1}{2ia}\int_0^{\infty} e^{-u}(-iu)^\frac{1-a}{a}\: (-i\: du) = \frac{(-i)^{\frac{1-a}{a}}}{2a}\int_0^{\infty} e^{-u}u^{\frac{1}{a}-1}\: du = \frac{(-i)^{\frac{1-a}{a}}}{2a}\Gamma\left(\frac{1}{a}\right) = \overline{I_1}$.

    So finally $\displaystyle I=I_1+I_2=I_1+\overline{I_1} = 2\Re(I_1)=\frac{\Gamma\left(\frac{1}{a}\right)\Re( i^{\frac{1-a}{a}})}{a}=\frac{\Gamma\left(\frac{1}{a}\right)\c os(\frac{\pi}{2}\frac{1-a}{a})}{a}=\frac{\Gamma\left(\frac{1}{a}\right)\si n(\frac{\pi}{2a})}{a}$
    Last edited by Bruno J.; Aug 23rd 2009 at 05:40 PM.
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  8. #53
    MHF Contributor Bruno J.'s Avatar
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    Here is my integral! I posted a particular case of this one not very long ago but here it is anyways because I think it's very nice.

    Determine for which values of $\displaystyle z$ (possibly complex) the following integral exists, and evaluate it for those $\displaystyle z$:

    $\displaystyle I(z)=\int_0^1\frac{\log(x)^z\log(1-x)}{x}\ dx$
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  9. #54
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    Quote Originally Posted by Bruno J. View Post
    Here is my integral! I posted a particular case of this one not very long ago but here it is anyways because I think it's very nice.

    Determine for which values of $\displaystyle z$ (possibly complex) the following integral exists, and evaluate it for those $\displaystyle z$:
    $\displaystyle I(z)=\int_0^1\frac{\log(x)^z\log(1-x)}{x}\ dx$
    $\displaystyle
    I(z)=\int_0^1\frac{\log(x)^z\log(1-x)}{x}\ dx=
    $

    $\displaystyle
    \int_0^\infty (-u)^z\ln(1-e^{-u})\;du=
    $ (Note: Let $\displaystyle x=e^{-u} $)

    $\displaystyle
    (-1)^{z+1}\int_0^\infty u^z\sum_{n=1}^{\infty} \frac{e^{-nu}}{n} \; du=
    $

    $\displaystyle
    (-1)^{z+1}\sum_{n=1}^{\infty} \frac{1}{n} \int_0^\infty u^z e^{-nu} \; du=
    $

    $\displaystyle
    (-1)^{z+1}\sum_{n=1}^{\infty} \frac{1}{n^{z+2}} \int_0^\infty t^z e^{-t} \; dt=
    $ (Note: Let $\displaystyle t=nu $)

    $\displaystyle
    (-1)^{z+1}\Gamma(z+1)\sum_{n=1}^{\infty} \frac{1}{n^{z+2}} $

    $\displaystyle
    (-1)^{z+1}\Gamma(z+1)\;\zeta(z+2)
    $

    which requires Re$\displaystyle (z+2)>1$, i.e. Re$\displaystyle (z)>-1$ to exist.
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  10. #55
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    Quote Originally Posted by Bruno J. View Post
    I attack $\displaystyle
    I=\int_0^{\infty} \sin (x^a) \ dx, \ \ a > 1.
    $

    First we substitute $\displaystyle x^a=t$. Then $\displaystyle ax^{a-1}dx=dt$, i.e. $\displaystyle dx = \frac{dt}{ax^{a-1}}=\frac{1}{a}{t^\frac{1-a}{a}}\: dt$.

    Now $\displaystyle \int_0^{\infty} \sin (x^a) \ dx = \frac{1}{a}\int_0^{\infty} \sin (t){t^\frac{1-a}{a}}\: dt = \frac{1}{2ia}\int_0^{\infty} (e^{it}-e^{-it}){t^\frac{1-a}{a}}\: dt$.

    Now consider $\displaystyle I_1=\frac{1}{2ia}\int_0^{\infty} e^{it}t^\frac{1-a}{a}\: dt$. Set $\displaystyle it=-u$, so $\displaystyle dt = i\: du$. Then the integral becomes

    $\displaystyle \frac{1}{2ia}\int_0^{\infty} e^{-u}(iu)^\frac{1-a}{a}\: i\: du = \frac{i^{\frac{1-a}{a}}}{2a}\int_0^{\infty} e^{-u}u^{\frac{1}{a}-1}\: du = \frac{i^{\frac{1-a}{a}}}{2a}\Gamma\left(\frac{1}{a}\right)$.

    Similarily, consider $\displaystyle I_2=\frac{-1}{2ia}\int_0^{\infty} e^{-it}t^\frac{1-a}{a}\: dt$. Set $\displaystyle it=u$, so $\displaystyle dt = -i\: du$. Then the integral becomes

    $\displaystyle \frac{-1}{2ia}\int_0^{\infty} e^{-u}(-iu)^\frac{1-a}{a}\: (-i\: du) = \frac{(-i)^{\frac{1-a}{a}}}{2a}\int_0^{\infty} e^{-u}u^{\frac{1}{a}-1}\: du = \frac{(-i)^{\frac{1-a}{a}}}{2a}\Gamma\left(\frac{1}{a}\right) = \overline{I_1}$.

    So finally $\displaystyle I=I_1+I_2=I_1+\overline{I_1} = 2\Re(I_1)=\frac{\Gamma\left(\frac{1}{a}\right)\Re( i^{\frac{1-a}{a}})}{a}=\frac{\Gamma\left(\frac{1}{a}\right)\c os(\frac{\pi}{2}\frac{1-a}{a})}{a}$
    which is also
    $\displaystyle
    \frac{\Gamma\left(\frac{1}{a}\right)\sin\frac{\pi} {2a}}{a}$
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  11. #56
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    Quote Originally Posted by NonCommAlg View Post
    3) $\displaystyle \int_0^{\infty}\cot^{-1}(ax) \cot^{-1}(bx) \ dx, \ \ a > 0, \ b > 0$

    4) $\displaystyle \int_0^{\pi} (\sin x)^a \sin(ax) \ dx, \ \ a > -1$
    no.3 the answer is $\displaystyle \frac{\pi}{2}(\frac{1}{a}+\frac{1}{b})\ln(a+b) - \frac{{\pi}}{2}(\frac{\ln{b}}{a} + \frac{\ln{a}}{b})$

    no.4 $\displaystyle \int_0^{\pi} e^{iax} (\sin(x))^a ~dx = \frac{\pi}{2^a}e^{i\frac{\pi}{2}a}$
    Last edited by simplependulum; Aug 21st 2009 at 07:49 PM.
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  12. #57
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    Quote Originally Posted by luobo View Post
    which is also
    $\displaystyle
    \frac{\Gamma\left(\frac{1}{a}\right)\sin\frac{\pi} {2a}}{a}$
    When $\displaystyle a = 2 $ , the integral becomes the Frensel Integral and the result is $\displaystyle \sqrt{\pi}/\sqrt{8}$


    yes you are right , it is $\displaystyle \frac{\sqrt{\pi}}{\sqrt{8}} $
    Last edited by simplependulum; Aug 21st 2009 at 07:50 PM.
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  13. #58
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    Quote Originally Posted by simplependulum View Post
    When $\displaystyle a = 2 $ , the integral becomes the Frensel Integral and the result is $\displaystyle \sqrt{\pi}/\sqrt{2}$
    Fresnel integrals: $\displaystyle \frac{\sqrt{\pi}}{2\sqrt{2}}$
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  14. #59
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    Quote Originally Posted by simplependulum View Post
    no.3 the answer is $\displaystyle \frac{\pi}{2}(\frac{1}{a}+\frac{1}{b})\ln(a+b) - \frac{{\pi}}{2}(\frac{\ln{b}}{a} + \frac{\ln{a}}{b})$
    correct!


    no.4 $\displaystyle \int_0^{\pi} e^{iax} (\sin(x))^a ~dx = \frac{\pi}{2^a}e^{i\frac{\pi}{2}a}$
    and that gives us: $\displaystyle \int_0^{\pi} (\sin x)^a \sin(ax) \ dx = \frac{\pi}{2^a} \sin \left(\frac{\pi a}{2} \right),$ which is the correct answer. i hope you'll also post your solutions!

    ---------------------------------------------

    Ok, this problem is nice:

    Problem: Suppose $\displaystyle f,g : [a,b] \longrightarrow (0, \infty)$ are continuous, $\displaystyle f$ is decreasing and $\displaystyle g$ is increasing. Prove that for all real numbers $\displaystyle c \geq d > 0: \ \ \frac{\int_a^b f^c(x) \ dx}{\int_a^b f^d(x) \ dx} \geq \frac{\int_a^b g(x)f^c(x) \ dx}{\int_a^b g(x) f^d(x) \ dx}.$

    Note: Here by $\displaystyle f^r(x)$ we mean $\displaystyle (f(x))^r.$ The solution I know doesn't use any known integral inequalities. Instead, it uses a nice integration technique and it's fairly short!
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  15. #60
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    Quote Originally Posted by NonCommAlg View Post
    Problem: Suppose $\displaystyle f,g : [a,b] \longrightarrow (0, \infty)$ are continuous, $\displaystyle f$ is decreasing and $\displaystyle g$ is increasing. Prove that for all real numbers $\displaystyle c \geq d > 0: \ \ \frac{\int_a^b f^c(x) \ dx}{\int_a^b f^d(x) \ dx} \geq \frac{\int_a^b g(x)f^c(x) \ dx}{\int_a^b g(x) f^d(x) \ dx}.$
    Since $\displaystyle c\geq d$ we know that $\displaystyle \tfrac{f^d(x)}{f^c(x)}$ must be non-decreasing - because f is non-increasing- (we can divide by $\displaystyle f$ since it is non-zero in our interval), thus we must have $\displaystyle
    \left( {\tfrac{{f^d \left( x \right)}}
    {{f^c \left( x \right)}} - \tfrac{{f^d \left( y \right)}}
    {{f^c \left( y \right)}}} \right) \cdot \left( {g\left( x \right) - g\left( y \right)} \right) \geqslant 0
    $

    Expand: $\displaystyle
    \tfrac{{f^d \left( x \right)}}
    {{f^c \left( x \right)}} \cdot g\left( x \right) + \tfrac{{f^d \left( y \right)}}
    {{f^c \left( y \right)}} \cdot g\left( y \right) \geqslant \tfrac{{f^d \left( x \right)}}
    {{f^c \left( x \right)}} \cdot g\left( y \right) + \tfrac{{f^d \left( y \right)}}
    {{f^c \left( y \right)}} \cdot g\left( x \right)
    $ the inequality will be preserved if we multiply by: $\displaystyle
    f^c \left( x \right) \cdot f^c \left( y\right) > 0
    $

    Thus: $\displaystyle
    f^d \left( x \right) \cdot g\left( x \right) \cdot f^c \left( y \right) + f^d \left( y \right) \cdot g\left( y \right) \cdot f^c \left( x \right) \geqslant f^c \left( y \right) \cdot g\left( y \right) \cdot f^d \left( x \right) + f^c \left( x \right) \cdot g\left( x \right) \cdot f^d \left( y \right)
    $

    Now integrate in $\displaystyle
    \left[ {a,b} \right] \times \left[ {a,b} \right]
    $ and the inequality follows since $\displaystyle g(x),f(x)>0$ in $\displaystyle \left[ {a,b} \right]$ -and so the integrals are positive, and we can divide-
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