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Thread: Integrals

  1. #1
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    Integrals

    Evaluate

    $\displaystyle \int_0^\infty\frac{1-\cos x}{x^2}\,dx$

    $\displaystyle \int_0^1\frac1{(x^2-x+1)(e^{2x-1}+1)}\,dx$
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  2. #2
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    Quote Originally Posted by liyi View Post
    Evaluate
    $\displaystyle \int_0^\infty\frac{1-\cos x}{x^2}\,dx$
    $\displaystyle 1 - \cos x = 2\sin^2 \frac{x}{2}$
    Thus,
    $\displaystyle 2\int_0^{\infty} \frac{\sin^2 \frac{x}{2}}{x^2} dx$
    Let $\displaystyle t=x/2$,
    $\displaystyle 2\int_0^{\infty} \frac{\sin^2 t}{4t^2} \cdot 2dt = \frac{\pi}{2}$
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  3. #3
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    Quote Originally Posted by liyi View Post
    Evaluate

    $\displaystyle \int_0^\infty\frac{1-\cos x}{x^2}\,dx$
    Solution 2. In my other post I used that fact that $\displaystyle \int_0^{\infty} \frac{\sin^2 t}{t^2}~dt = \frac{\pi}{2}$ (it was proven on this forum before).

    Here is a method just using the fact that $\displaystyle \int_0^{\infty} \frac{\sin x}{x} dx = \frac{\pi}{2}$.

    Note that,
    $\displaystyle \int_0^{\infty} \frac{1-\cos x}{x^2} dx = \int_0^{\infty} \int_0^y \frac{\sin y}{x^2} dy~dx = $$\displaystyle \int_0^{\infty} \int_y^{\infty} \frac{\sin y}{x^2} dx~dy = \int_0^{\infty} \frac{\sin y}{y} dy = \frac{\pi}{2}$
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  4. #4
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    Quote Originally Posted by liyi View Post
    Evaluate

    $\displaystyle \int_0^\infty\frac{1-\cos x}{x^2}\,dx$
    Here's another proof.

    We'll use the following fact $\displaystyle \int_0^\infty\frac{\sin x}x\,dx=\frac\pi2.$ (Proven here.)

    Now $\displaystyle \frac{{1 - \cos x}}
    {x} = \int_0^1 {\sin (ux)\,du} .$

    The integral becomes to $\displaystyle \int_0^\infty {\int_0^1 {\frac{{\sin (ux)}}
    {x}\,du} \,dx} = \int_0^1 {\int_0^\infty {\frac{{\sin (ux)}}
    {x}\,dx} \,du} .$

    For the inner integral, substitute $\displaystyle \varphi=ux,$

    $\displaystyle \int_0^\infty {\frac{{\sin (ux)}}
    {x}\,dx} = \frac{1}
    {u}\int_0^\infty {\frac{{u\sin \varphi }}
    {\varphi }\,d\varphi } = \frac{\pi }
    {2}.$

    Finally

    $\displaystyle \int_0^\infty {\frac{{1 - \cos x}}
    {{x^2 }}\,dx} = \frac{\pi }
    {2}\int_0^1 {du} = \frac{\pi }
    {2}\,\blacksquare$

    --

    A related problem.

    --

    As for $\displaystyle \int_0^\infty {\frac{{\sin ^2 x}}
    {{x^2 }}\,dx}$ can be found here.
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  5. #5
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    Nice problem

    Quote Originally Posted by liyi View Post
    $\displaystyle \int_0^1\frac1{(x^2-x+1)(e^{2x-1}+1)}\,dx$
    Define $\displaystyle u=2x-1,$

    $\displaystyle \int_0^1 {\frac{1}
    {{\left( {x^2 - x + 1} \right)\left( {e^{2x - 1} + 1} \right)}}\,dx} = \int_{ - 1}^1 {\frac{2}
    {{\left( {u^2 + 3} \right)\left( {e^u + 1} \right)}}\,du} .$

    Substitute $\displaystyle u=-\varphi,$

    $\displaystyle \int_{ - 1}^1 {\frac{2}
    {{\left( {u^2 + 3} \right)\left( {e^u + 1} \right)}}\,du} = \int_{ - 1}^1 {\frac{2}
    {{\left( {\varphi ^2 + 3} \right)\left( {e^{ - \varphi } + 1} \right)}}\,d\varphi ,}$ so

    $\displaystyle \tau = \int_{ - 1}^1 {\frac{2}
    {{\left( {u^2 + 3} \right)\left( {e^u + 1} \right)}}\,du} = \int_{ - 1}^1 {\frac{2}
    {{\left( {u^2 + 3} \right)\left( {e^{ - u} + 1} \right)}}\,du} .$

    And finally

    $\displaystyle 2\tau = \int_{ - 1}^1 {\frac{2}
    {{u^2 + 3}}\,du} = \frac{{2\pi }}
    {{3\sqrt 3 }}\,\therefore \,\tau = \frac{\pi }
    {{3\sqrt 3 }}.$
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