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    (Original post by Mladenov)
    Problem 210**

    Find \displaystyle \int_{0}^{1} \frac{(1-x^{\alpha})(1-x^{\beta})}{(1-x)\ln x} \, dx, for \alpha, \beta >-1 and \alpha+\beta >-1.
    I resist coming on for a few days and miss some definite integrals! fml :rolleyes:

    I'm sure there is a much nicer solution that takes advantages of the symmetries...

    Solution 210

    Let \displaystyle \phi(\alpha)=\int_{0}^{1} \frac{(1-x^{\alpha})(1-x^{\beta})}{(1-x)\ln x} \, dx,

    We note that \displaystyle \phi(0)=0,

    \displaystyle \begin{aligned} \Righatarrow \frac{\partial}{\partial \alpha} \phi(\alpha) =-\int_0^1 \frac{x^{\alpha} (1-x^{\beta})}{(1-x)} dx = -\int_0^1 x^{\alpha} \sum_{n=1}^{\beta} x^{n-1} \ dx = -\sum_{n=1}^{\beta} \int_0^1 x^{\alpha + n-1} \ dx \end{aligned}

\displaystyle =-\sum_{n=1}^{\beta} \left[\frac{x^{\alpha +n}}{\alpha +n} \right]_0^1 = -\sum_{n=1}^{\beta} \frac{1}{\alpha +n}

    Here we have used the given conditions in assuring that there exists no value of n for which of of the terms in this series would have integrated to form a logarithm.

    \displaystyle \Rightarrow \phi(\alpha) = -\int_0^{\alpha} \sum_{n=1}^{\beta} \frac{1}{t +n} \ dt = \sum_{n=1}^{\beta} \left[ \ln(t+n) \right]_{\alpha}^0 = \sum_{n=1}^{\beta} \ln(\frac{n}{\alpha +n}) 

\displaystyle = \ln \left(\frac{\alpha ! \beta !}{(\alpha +\beta)!} \right) = -\ln \binom{\alpha+\beta}{\alpha}\ \square
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    (Original post by Mladenov)
    Problem 209*

    Evaluate \displaystyle \int_{0}^{\infty} \frac{A_{1}\cos a_{1}x + \cdots + A_{k}\cos a_{k}x}{x}dx when a_{i}>0 for all i \in \{1,2, \cdots,k \} and A_{1}+\cdots+A_{k}=0.
    Not sure this is right...

    Let \displaystyle f(t)=\int_0^{\infty} \frac{\cos(tx)}{x} \ dx,

    \displaystyle \begin{aligned} \Rightarrow \mathcal{L} \{ f(t) \} = \int_0^{\infty} \frac{1}{x} \int_0^{\infty} e^{-st} \cos(tx) \ dt \dx = \int_0^{\infty} \frac{s}{x(x^2+s^2)} \ dx = \frac{1}{s} \left[ \ln \left( \frac{x}{ \sqrt{x^2+s^2}} \right) \right]_0^{\infty} \end{aligned}

    Let \displaystyle I(x) = \int_{0}^{\infty} \frac{A_{1}\cos a_{1}x + \cdots + A_{k}\cos a_{k}x}{x}dx = \sum_{n=1}^k A_n f(a_n),

    \displaystyle \begin{aligned} \Rightarrow \mathcal{L} \{ I(x) \} = \frac{1}{s} \sum_{n=1}^k A_n \left[ \ln \left( \frac{x}{\sqrt{x^2+s^2}} \right) \right]_0^{\infty} = \frac{1}{s} \left[ \sum_{n=1}^k \ln \left( \frac{x}{\sqrt{x^2+s^2}} \right)^{A_n} \right]_0^{\infty} \end{aligned}

\displaystyle = \frac{1}{s} \left[ \ln \left( \frac{x}{\sqrt{x^2+s^2}} \right)^{A_1+A_2+ \cdots + A_n} \right]_0^{\infty}= \frac{1}{s} \left[ \ln(1) \right]_0^{\infty} =0

    \displaystyle \Rightarrow I(x) = \mathcal{L}^{-1} \{0 \} = 0 \ \square
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    (Original post by Jkn)
    Not sure this is right...

    Solution 209

    Let \displaystyle f(t)=\int_0^{\infty} \frac{\cos(tx)}{x} \ dx,

    \displaystyle \begin{aligned} \Rightarrow \mathcal{L} \{ f(t) \} = \int_0^{\infty} \frac{1}{x} \int_0^{\infty} e^{-st} \cos(tx) \ dt \dx = \int_0^{\infty} \frac{s}{x(x^2+s^2)} \ dx = \frac{1}{s} \left[ \ln \left( \frac{x}{ \sqrt{x^2+s^2}} \right) \right]_0^{\infty} \end{aligned}

    Let \displaystyle I(x) = \int_{0}^{\infty} \frac{A_{1}\cos a_{1}x + \cdots + A_{k}\cos a_{k}x}{x}dx = \sum_{n=1}^k A_n f(a_n),

    \displaystyle \begin{aligned} \Rightarrow \mathcal{L} \{ I(x) \} = \frac{1}{s} \sum_{n=1}^k A_n \left[ \ln \left( \frac{x}{\sqrt{x^2+s^2}} \right) \right]_0^{\infty} = \frac{1}{s} \left[ \sum_{n=1}^k \ln \left( \frac{x}{\sqrt{x^2+s^2}} \right)^{A_n} \right]_0^{\infty} \end{aligned}

\displaystyle = \frac{1}{s} \left[ \ln \left( \frac{x}{\sqrt{x^2+s^2}} \right)^{A_1+A_2+ \cdots + A_n} \right]_0^{\infty}= \frac{1}{s} \left[ \ln(1) \right]_0^{\infty} =0

    \displaystyle \Rightarrow I(x) = \mathcal{L}^{-1} \{0 \} = 0 \ \square
    You invoked \displaystyle \mathcal{L} \{ I(x) \}\left(s\right) = \sum_{n=1}^k A_n \mathcal{L} \left\{ f(a_n) \right\} \left(s\right), but instead of \mathcal{L} \left\{ f(a_n) \right\}\left(s\right) you subbed in the expression for \mathcal{L} \left\{ f(t) \right\}\left(s\right) instead, which is not equivalent in general (and in any case f(a_n) is a constant)....
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    (Original post by ukdragon37)
    You invoked \displaystyle \mathcal{L} \{ I(x) \}\left(s\right) = \sum_{n=1}^k A_n \mathcal{L} \left\{ f(a_n) \right\} \left(s\right), but instead of \mathcal{L} \left\{ f(a_n) \right\}\left(s\right) you subbed in the expression for \mathcal{L} \left\{ f(t) \right\}\left(s\right) instead, which is not equivalent in general (and in any case f(a_n) is a constant)....
    But, given the fact I have shown that \mathcal{L} \{ f(t) \} is independent of t, would that not be sufficient? :/ I mean, surely the laplace transform of f(a) is the same as f(b)? :/
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    (Original post by Mladenov)

    Problem 209*

    Evaluate \displaystyle \int_{0}^{\infty} \frac{A_{1}\cos a_{1}x + \cdots + A_{k}\cos a_{k}x}{x}dx when a_{i}>0 for all i \in \{1,2, \cdots,k \} and A_{1}+\cdots+A_{k}=0.
    .
    I enjoyed this one.

    Solution 209

    Observe that,

     \beginaligned \displaystyle  \frac{ \cos {a_1 x} - \cos{a_2 x}}{x} =  \displaystyle \int_{a_2}^{a_1} - \sin {ax}\ da = Im \left[ \displaystyle \int_{a_2}^{a_1} e^{-iax}\ da \right]



\Rightarrow \displaystyle \int_0^{\infty}  \displaystyle  \frac{ \cos {a_1 x} - \cos{a_2 x}}{x}\ dx =  \displaystyle \int_0^{\infty}  Im \left[ \displaystyle \int_{a_2}^{a_1} e^{-iax}\ da \right]\ dx



\beginaligned = \displaystyle \int_{a_2}^{a_1} Im \displaystyle \int_0^{\infty} \left[  e^{-iax} \right]\ dx\ da





\beginaligned = \displaystyle \int_{a_2}^{a_1} Im \left[ - \frac{e^{-iax}}{ia} \right] _0^{\infty}\ da





\beginaligned =  \displaystyle \int_{a_2}^{a_1} Im \left[ -\frac{i}{a} \right]\ da





\beginaligned = \displaystyle \int_{a_2}^{a_1} -\frac{1}{a}\ da





\beginaligned = \displaystyle \ln { \frac{a_2}{a_1}}

    Now, notice that,  A_k = -A_1 -A_2 \cdots

    So,

     \displaystyle \int_{0}^{\infty} \frac{A_{1}\cos a_{1}x + \cdots + A_{k}\cos a_{k}x}{x}dx



\beginaligned =  A_1 \displaystyle \int_0^{\infty}  \displaystyle  \frac{ \cos {a_1 x} - \cos{a_k x}}{x}\ dx + \cdots A_{k-1}\displaystyle \int_0^{\infty}  \displaystyle  \frac{ \cos {a_{k-1} x} - \cos{a_k x}}{x}\ dx

    



\beginaligned = A_1 \ln {\frac{a_k}{a_1}} + \cdots A_{k-1} \ln {\frac{a_k}{a_{k-1}}}

    



\beginaligned = \ln {\frac{a_k^{A_1+A_2 + \cdots}}{a_1^{A_1} a_2^{A_2} \cdots}}

    



\beginaligned =  \ln {\frac{a_k^{-A_k}}{a_1^{A_1} a_2^{A_2} \cdots}}

    



\beginaligned = - \ln {(a_1^{A_1} a_2^{A_2} \cdots a_k^{A_k})}
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    (Original post by Jkn)
    But, given the fact I have shown that \mathcal{L} \{ f(t) \} is independent of t, would that not be sufficient? :/ I mean, surely the laplace transform of f(a) is the same as f(b)? :/
    But that's true for all Laplace transforms - instead of being dependent on t (since you essentially integrate over all values of t) it becomes dependent on s instead. However if you a priori substitute t with one particular value, that ends up becoming a different transform (in particular since f\left(a_n\right) is a constant we have \mathcal{L} \{ f(a_n) \}(s) = \dfrac{f(a_n)}{s}).
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    (Original post by ukdragon37)
    But that's true for all Laplace transforms - instead of being dependent on t (since you essentially integrate over all values of t) it becomes dependent on s instead. However if you a priori substitute t with one particular value, that ends up becoming a different transform (in particular since f\left(a_n\right) is a constant we have \mathcal{L} \{ f(a_n) \}(s) = \dfrac{f(a_n)}{s}).
    Hmm, perhaps I am doing stupid things with a_i :/

    What is the flaw in this argument: (?)

    Let  x \to \frac{x}{a_i},

    \displaystyle \Rightarrow \int_0^{\infty} \frac{\cos(a_i x)}{x} dx = \int_0^{\infty} \frac{\cos(x)}{x} dx

    for all a_i.

    \displaystyle \Rightarrow I(x) = \int_0^{\infty} (A_1+A_2+...+A_n) \frac{\cos(x)}{x} dx = 0

    :/ ??

    (Original post by MW24595)
    Spoiler:
    Show
    I enjoyed this one.

    Solution 209

    Observe that,

     \beginaligned \displaystyle  \frac{ \cos {a_1 x} - \cos{a_2 x}}{x} =  \displaystyle \int_{a_2}^{a_1} - \sin {ax}\ da = Im \left[ \displaystyle \int_{a_2}^{a_1} e^{-iax}\ da \right]



\Rightarrow \displaystyle \int_0^{\infty}  \displaystyle  \frac{ \cos {a_1 x} - \cos{a_2 x}}{x}\ dx =  \displaystyle \int_0^{\infty}  Im \left[ \displaystyle \int_{a_2}^{a_1} e^{-iax}\ da \right]\ dx



\beginaligned = \displaystyle \int_{a_2}^{a_1} Im \displaystyle \int_0^{\infty} \left[  e^{-iax} \right]\ dx\ da





\beginaligned = \displaystyle \int_{a_2}^{a_1} Im \left[  e^{-iax} \right] _0^{\infty}\ da





\beginaligned =  \displaystyle \int_{a_2}^{a_1} Im \left[ -\frac{i}{a} \right]\ da





\beginaligned = \displaystyle \int_{a_2}^{a_1} -\frac{1}{a}\ da





\beginaligned = \displaystyle \ln { \frac{a_2}{a_1}}

    Now, notice that,  A_k = -A_1 -A_2 \cdots

    So,

     \displaystyle \int_{0}^{\infty} \frac{A_{1}\cos a_{1}x + \cdots + A_{k}\cos a_{k}x}{x}dx



\beginaligned =  A_1 \displaystyle \int_0^{\infty}  \displaystyle  \frac{ \cos {a_1 x} - \cos{a_k x}}{x}\ dx + \cdots A_{k-1}\displaystyle \int_0^{\infty}  \displaystyle  \frac{ \cos {a_{k-1} x} - \cos{a_k x}}{x}\ dx

    



\beginaligned = A_1 \ln {\frac{a_k}{a_1}} + \cdots A_{k-1} \ln {\frac{a_k}{a_{k-1}}}

    



\beginaligned = \ln {\frac{a_k^{A_1+A_2 + \cdots}}{a_1^{A_1} a_2^{A_2} \cdots}}

    



\beginaligned =  \ln {\frac{a_k^{-A_k}}{a_1^{A_1} a_2^{A_2} \cdots}}

    



\beginaligned = - \ln {(a_1^{A_1} a_2^{A_2} \cdots a_k^{A_k})}
    Nice solutions bro!
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    (Original post by Jkn)
    Hmm, perhaps I am doing stupid things with a_i :/

    What is the flaw in this argument: (?)

    Let  x \to \frac{x}{a_i},

    \displaystyle \Rightarrow \int_0^{\infty} \frac{\cos(a_i x)}{x} dx = \int_0^{\infty} \frac{\cos(x)}{x} dx

    for all a_i.

    \displaystyle \Rightarrow I(x) = \int_0^{\infty} (A_1+A_2+...+A_n) \frac{\cos(x)}{x} dx = 0

    :/ ??
    You are indeed being clumsy with the a_is Each substitution of x \to \frac{x}{a_i} gives rise to a different but related x, while you are presuming they become the same x. Instead you should be careful and write:

    Let x \to \dfrac{x_i}{a_i}


    \displaystyle \Rightarrow \int_0^{\infty} \frac{\cos(a_i x)}{x} dx = \int_0^{\infty} \frac{\cos(x_i)}{x_i} dx_i


    \displaystyle \Rightarrow I(x) = \sum_{i=1}^k A_i \int_0^{\infty} \frac{\cos(x_i)}{x_i} dx_i

    EDIT: oh I see what you mean, you are asking why each \int_0^{\infty} \frac{\cos(x_i)}{x_i} dx_i doesn't give the same result. Well, that's because this method suffers from the more fundamental problem that each \int_0^{\infty} \frac{\cos(x_i)}{x_i} dx_i actually does not converge and an improper integral of sums can only be split into a sum of improper integrals (analogous to the sum of limits vs. limit of sums) only if the component limits exist. In such a case you are not entitled to use a substitution which applies differently to each component of the sum, only one which applies to the integral as a whole.
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    (Original post by ukdragon37)
    You are indeed being clumsy with the a_is Each substitution of x \to \frac{x}{a_i} gives rise to a different but related x, while you are presuming they become the same x. Instead you should be careful and write:

    Let x \to \dfrac{x_i}{a_i}


    \displaystyle \Rightarrow \int_0^{\infty} \frac{\cos(a_i x)}{x} dx = \int_0^{\infty} \frac{\cos(x_i)}{x_i} dx_i


    \displaystyle \Rightarrow I(x) = \sum_{i=1}^k A_i \int_0^{\infty} \frac{\cos(x_i)}{x_i} dx_i

    EDIT: oh I see what you mean, you are asking why each \int_0^{\infty} \frac{\cos(x_i)}{x_i} dx_i doesn't give the same result. Well, that's because this method suffers from the more fundamental problem that each \int_0^{\infty} \frac{\cos(x_i)}{x_i} dx_i actually does not converge and an improper integral of sums can only be split into a sum of improper integrals (analogous to the sum of limits vs. limit of sums) only if the component limits exist. In such a case you are not entitled to use a substitution which applies differently to each component of the sum, only one which applies to the integral as a whole.
    Ah thanks for explaining that! Is that similar to the kind of mistakes you can make when dividing by zero? ...or if you try and express something in partial fractions and incorrectly assume the numerators of quadratic terms are always constant? i.e. incorrectly assuming something exists and, hence, losing a great deal of generality in your solution?

    Mathematics is full of surprises... I suppose I need to learn analysis before I use such sophisticated methods :rollseyes:
    (Original post by Mladenov)
    Edgar Allan Poe is an influential writer, you know.
    A fellow mathematician who likes Poe! I never thought I would see the day!

    "Deep into that darkness peering, long I stood there, wondering, fearing, doubting, dreaming dreams no mortal ever dared to dream before."
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    (Original post by Jkn)
    Ah thanks for explaining that! Is that similar to the kind of mistakes you can make when dividing by zero? ...or if you try and express something in partial fractions and incorrectly assume the numerators of quadratic terms are always constant? i.e. incorrectly assuming something exists and, hence, losing a great deal of generality in your solution?

    Mathematics is full of surprises... I suppose I need to learn analysis before I use such sophisticated methods :rollseyes:
    Well yes, they all fall into the general category of "assuming something that's normally safe to assume but actually in this case is false".
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    (Original post by Jkn)

    Nice solutions bro!
    Lol, thanks, man.

    How're the exams coming along?
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    (Original post by ukdragon37)
    Well yes, they all fall into the general category of "assuming something that's normally safe to assume but actually in this case is false".
    Is this correct: (?)

    \displaystyle \begin{aligned} \int_a^b \{ f(x) + g(x) \} dx = \int_a^b f(x) dx + \int_a^b g(x) dx \iff \left( \int_a^b f(x) dx , \int_a^b g(x) dx \right) \in \mathbb{R} \end{aligned}
    (Original post by MW24595)
    Lol, thanks, man.

    How're the exams coming along?
    Where have you developed such a confident use of complex numbers in evaluating integrals? Has this just been though practice, exploration and/or further reading or have you been formally taught it in school?

    Not too badly so far! Manage to get through my English and Spanish without any disasters (no idea what grades I have though..) which is good considering I had no time to revise either!

    Also had by D2 exam this morning, though, for obvious reasons, cannot comment.

    Have you been doing any exams this summer?
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    (Original post by Jkn)
    Is this correct: (?)

    \displaystyle \begin{aligned} \int_a^b \{ f(x) + g(x) \} dx = \int_a^b f(x) dx + \int_a^b g(x) dx \iff \left( \int_a^b f(x) dx , \int_a^b g(x) dx \right) \in \mathbb{R} \end{aligned}
    No, there is no reason why they have to be in \mathbb R - it works for definite integrals that end up being complex values too.
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    (Original post by ukdragon37)
    No, there is no reason why they have to be in \mathbb R - it works for definite integrals that end up being complex values too.
    So

     \displaystyle\left( \int_a^b f(x) dx , \int_a^b g(x) dx \right)\in \mathbb{R} \implies  \begin{aligned} \int_a^b \{ f(x) + g(x) \} dx = \int_a^b f(x) dx + \int_a^b g(x) dx 

\end{aligned}
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    (Original post by bananarama2)
     \displaystyle\left( \int_a^b f(x) dx , \int_a^b g(x) dx \right)\in \mathbb{R}
    What does that even mean?

    I think all people who are taking STEP should now focus their attention only on it.
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    (Original post by jack.hadamard)
    What does that even mean?

    I think all people who are taking STEP should now focus their attention only on it.
    I was hesitant in giving an affirmative answer because I was thinking if I thought really hard (or if I even have the time, dissertation not withstanding, despite that linearity usually holds) I might be able to come up with a counterexample perhaps by letting a and b be non-real or improper.
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    (Original post by jack.hadamard)
    What does that even mean?

    I think all people who are taking STEP should now focus their attention only on it.
    No idea, I was just copying, pasting and hoping for the best. You're right, but I'm not doing STEP
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    (Original post by Jkn)
    Where have you developed such a confident use of complex numbers in evaluating integrals? Has this just been though practice, exploration and/or further reading or have you been formally taught it in school?

    Not too badly so far! Manage to get through my English and Spanish without any disasters (no idea what grades I have though..) which is good considering I had no time to revise either!

    Also had by D2 exam this morning, though, for obvious reasons, cannot comment.

    Have you been doing any exams this summer?
    Lol, Just through exploration, I guess.
    I've always liked Complex Numbers, so, even while computing bromidic integrals, I try complex-ifying it sometimes to see what happens.


    Nice.

    Lol, D2, eh? How come?
    Gah, I know nothing of it. We do Statistics and Mechanics over here.

    Nope, no exams. The upcoming summer (monsoon out here, really) is one that seems sodden with gratuitous vacuity. It's good because I can do what I like, but not good, because I've no motivation to really do anything. :lol:
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    (Original post by ukdragon37)
    No, there is no reason why they have to be in \mathbb R - it works for definite integrals that end up being complex values too.
    Oh yeah, that's what I meant! :lol: What I mean to say is: Can the theorem be applied only when each definite integral has a finite value?
    (Original post by jack.hadamard)
    What does that even mean?
    That each definite integral takes a value that is in that set! Don't know why everyone's always outraged with use of what is probably bad notation :lol:

    At A-Level we obviously aren't taught about set theory but we are told we can use \in to say something is part of a set of numbers. What's so wrong about what I wrote that means you cannot understand it?
    I think all people who are taking STEP should now focus their attention only on it.
    Why?
    (Original post by MW24595)
    Lol, Just through exploration, I guess.
    I've always liked Complex Numbers, so, even while computing bromidic integrals, I try complex-ifying it sometimes to see what happens.

    Nice.

    Lol, D2, eh? How come?
    Gah, I know nothing of it. We do Statistics and Mechanics over here.

    Nope, no exams. The upcoming summer (monsoon out here, really) is one that seems sodden with gratuitous vacuity. It's good because I can do what I like, but not good, because I've no motivation to really do anything. :lol:
    Oh awesome! BROmidic integrals? Has that got something to do with being a bro? :lol:

    Oh it sucks! I had to do it for Additional Further Maths Doing all 18 modules!

    Oh right hmm, why didn't you just go to university last year?
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    (Original post by Jkn)
    A fellow mathematician who likes Poe! I never thought I would see the day!

    "Deep into that darkness peering, long I stood there, wondering, fearing, doubting, dreaming dreams no mortal ever dared to dream before."
    Actually, apart from Poe, Ayn Rand is the only writer whom books I can enjoy.

    By the way, Jkn, do you have an elegant, and not arduous, solution to problem 198, for I have one which is a hell of an arithmetic and I am not quite enthusiastic to type it in latex.

    Problem 211**

    Evaluate \displaystyle \int_{-\infty}^{\infty} \frac{\sinh ax}{\sinh bx} \, dx, where b> |a|.

    Problem 212**

    Evaluate \displaystyle \int_{0}^{\infty} \frac{2-2\cos x-x\sin x}{x^{4}}\, dx
 
 
 
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