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  1. MHB Apprentice

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    #1
    I am Trying to solve the difference of the two following integrals:

    (1) $g_{1}(x) = \int_{0}^{\infty} \frac{cos(kx)}{k}\,dk$
    (2) $g_{2}(x) = \int_{0}^{\infty} \frac{\exp(-2k)cos(kx)}{k}\,dk$

    I read the thread on Advanced Integration Techniques and it mentioned the Differentiation under the integral sign technique which I am unfamiliar to. Nonetheless I tried it and found ,
    $g_{1}(x)=\infty $ and $g_{2}(x)= \infty + \frac{1}{2}ln(1+\frac{4}{x^2})$

    Hence $g_{1}(x) - g_{2}(x) = -\frac{1}{2}ln(1+\frac{4}{x^2})$.

    What I would like to know is my final answer true or false and if it is false where did I make an invalid assumption or mistake?

  2. MHB Master
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    #2
    Quote Originally Posted by MBM View Post
    I am Trying to solve the difference of the two following integrals:

    (1) $g_{1}(x) = \int_{0}^{\infty} \frac{cos(kx)}{k}\,dk$
    (2) $g_{2}(x) = \int_{0}^{\infty} \frac{\exp(-2k)cos(kx)}{k}\,dk$

    I read the thread on Advanced Integration Techniques and it mentioned the Differentiation under the integral sign technique which I am unfamiliar to. Nonetheless I tried it and found ,
    $g_{1}(x)=\infty $ and $g_{2}(x)= \infty + \frac{1}{2}ln(1+\frac{4}{x^2})$

    Hence $g_{1}(x) - g_{2}(x) = -\frac{1}{2}ln(1+\frac{4}{x^2})$.

    What I would like to know is my final answer true or false and if it is false where did I make an invalid assumption or mistake?
    You got the difference $\infty - \infty$ and you set it equal to $0$, but this difference is undefined.


    Using Euler's formula $$e^{ix}=\cos{(x)}+i \sin{(x)}$$
    we get $$\cos{(x)}=\frac{e^{ix}+e^{-ix}}{2}$$

    Therefore we have the following:

    $$\int_{0}^{\infty} \frac{cos(kx)}{k}\,dk- \int_{0}^{\infty} \frac{e^{-2k}cos(kx)}{k}\,dk =\int_{0}^{\infty} \frac{e^{ikx}+e^{-ikx}}{2k}\,dk -\int_{0}^{\infty} \frac{e^{-2k}(e^{ikx}+e^{-ikx})}{2k}\,dk= \\ \int_{0}^{\infty} \frac{e^{ikx}+e^{-ikx}}{2k}\,dk -\int_{0}^{\infty} \frac{e^{(ix-2)k}+e^{-(ix+2)k}}{2k}\,dk$$

  3. MHB Apprentice

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    #3 Thread Author
    Quote Originally Posted by mathmari View Post
    You got the difference $\infty - \infty$ and you set it equal to $0$, but this difference is undefined.


    Using Euler's formula $$e^{ix}=\cos{(x)}+i \sin{(x)}$$
    we get $$\cos{(x)}=\frac{e^{ix}+e^{-ix}}{2}$$

    Therefore we have the following:

    $$\int_{0}^{\infty} \frac{cos(kx)}{k}\,dk- \int_{0}^{\infty} \frac{e^{-2k}cos(kx)}{k}\,dk =\int_{0}^{\infty} \frac{e^{ikx}+e^{-ikx}}{2k}\,dk -\int_{0}^{\infty} \frac{e^{-2k}(e^{ikx}+e^{-ikx})}{2k}\,dk= \\ \int_{0}^{\infty} \frac{e^{ikx}+e^{-ikx}}{2k}\,dk -\int_{0}^{\infty} \frac{e^{(ix-2)k}+e^{-(ix+2)k}}{2k}\,dk$$

    @ mathmari, thank you for your response. You have decomposed the integral in to 4 complex exponential integrals multiplied by a hyperbola however I do not see how it simplifies the integral, i.e. consider the following integral

    $\int_{0}^{\infty} \frac{\exp(ikx)}{2k}\,dk$ is still difficult to solve with the knowledge that I have. (or maybe I am not thinking)

  4. MHB Master
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    #4
    Quote Originally Posted by MBM View Post
    I am Trying to solve the difference of the two following integrals:

    (1) $g_{1}(x) = \int_{0}^{\infty} \frac{cos(kx)}{k}\,dk$
    (2) $g_{2}(x) = \int_{0}^{\infty} \frac{\exp(-2k)cos(kx)}{k}\,dk$

    I read the thread on Advanced Integration Techniques and it mentioned the Differentiation under the integral sign technique which I am unfamiliar to. Nonetheless I tried it and found ,
    $g_{1}(x)=\infty $ and $g_{2}(x)= \infty + \frac{1}{2}ln(1+\frac{4}{x^2})$

    Hence $g_{1}(x) - g_{2}(x) = -\frac{1}{2}ln(1+\frac{4}{x^2})$.

    What I would like to know is my final answer true or false and if it is false where did I make an invalid assumption or mistake?
    The problem is bad set because both integrals diverge. It is reasonable instead the calculation of the integral...

    $\displaystyle g(x) = \int_{0}^{\infty} \frac{1 - e^{-2\ k}}{k}\ \cos k\ x\ dk\ (1)$

    ... which converges. The calculation of (1) is realatively easy using some advanced technique like the Laplace Transform, much more difficult using an elementary approach...

    Kind regards

    $\chi$ $\sigma$
    Last edited by chisigma; September 4th, 2014 at 08:57.

  5. زيد اليافعي
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    #5
    $$F(y)=\int^\infty_0 \frac{\cos(x)(1-e^{-2xy})}{x} dx$$

    $$F'(y)=2\int^\infty_0 e^{-2xy}\cos(x)\,dx = \frac{4y}{1+4y^2}$$

    $$F(y)=\frac{\log(1+4y^2)}{2}$$

    Hence we have

    $$F\left(\frac{1}{y}\right)=\int^\infty_0 \frac{\cos(x)(1-e^{-2\frac{x}{y}})}{x} dx$$

    Now let $x/y=t$

    $$\int^\infty_0 \frac{\cos(yt)(1-e^{-2t})}{t} dt =\frac{\log\left(1+\frac{4}{y^2} \right)}{2} $$

  6. MHB Apprentice

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    #6 Thread Author
    Quote Originally Posted by ZaidAlyafey View Post
    $$F(y)=\int^\infty_0 \frac{\cos(x)(1-e^{-2xy})}{x} dx$$

    $$F'(y)=2\int^\infty_0 e^{-2xy}\cos(x)\,dx = \frac{4y}{1+4y^2}$$

    $$F(y)=\frac{\log(1+4y^2)}{2}$$

    Hence we have

    $$F\left(\frac{1}{y}\right)=\int^\infty_0 \frac{\cos(x)(1-e^{-2\frac{x}{y}})}{x} dx$$

    Now let $x/y=t$

    $$\int^\infty_0 \frac{\cos(yt)(1-e^{-2t})}{t} dt =\frac{\log\left(1+\frac{4}{y^2} \right)}{2} $$
    thanks a lot. how did you become good in math, do you have any books you could recommend for me to read?

  7. زيد اليافعي
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    #7
    Quote Originally Posted by MBM View Post
    thanks a lot. how did you become good in math, do you have any books you could recommend for me to read?
    You're welcome.

    Essentially, my knowledge (of this particular field) is based on reading different articles, papers and books. I would suggest some books but that depends on the area you're interested in.

    Since this is not the best place to discuss that , I prefer you present your concerns via visitor messages or private messages.

  8. MHB Master
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    #8
    Quote Originally Posted by ZaidAlyafey View Post
    $$F(y)=\int^\infty_0 \frac{\cos(x)(1-e^{-2xy})}{x} dx$$

    $$F'(y)=2\int^\infty_0 e^{-2xy}\cos(x)\,dx = \frac{4y}{1+4y^2}$$
    From the formal point of view there is some problem in this step since the rule of derivation under the integral sign is valid only if the range of integration is limited...



    In this particular case, it should be noted that setting y = 0 would lead to the conclusion that ...

    $\displaystyle F^{\ '} (0) = \int_{0}^{\infty} \cos x\ dx = 0\ (1)$

    which of course is not true...

    Kind regards

    $\chi$ $\sigma$

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