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Math Help - Integrals

  1. #1
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    Integrals

    Evaluate

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

    \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
    \int_0^\infty\frac{1-\cos x}{x^2}\,dx
    1 - \cos x = 2\sin^2 \frac{x}{2}
    Thus,
    2\int_0^{\infty} \frac{\sin^2 \frac{x}{2}}{x^2} dx
    Let t=x/2,
    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

    \int_0^\infty\frac{1-\cos x}{x^2}\,dx
    Solution 2. In my other post I used that fact that \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 \int_0^{\infty} \frac{\sin x}{x} dx = \frac{\pi}{2}.

    Note that,
    \int_0^{\infty} \frac{1-\cos x}{x^2} dx = \int_0^{\infty} \int_0^y \frac{\sin y}{x^2} dy~dx = \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

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

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

    Now \frac{{1 - \cos x}}<br />
{x} = \int_0^1 {\sin (ux)\,du} .

    The integral becomes to \int_0^\infty {\int_0^1 {\frac{{\sin (ux)}}<br />
{x}\,du} \,dx} = \int_0^1 {\int_0^\infty {\frac{{\sin (ux)}}<br />
{x}\,dx} \,du} .

    For the inner integral, substitute \varphi=ux,

    \int_0^\infty {\frac{{\sin (ux)}}<br />
{x}\,dx} = \frac{1}<br />
{u}\int_0^\infty {\frac{{u\sin \varphi }}<br />
{\varphi }\,d\varphi } = \frac{\pi }<br />
{2}.

    Finally

    \int_0^\infty {\frac{{1 - \cos x}}<br />
{{x^2 }}\,dx} = \frac{\pi }<br />
{2}\int_0^1 {du} = \frac{\pi }<br />
{2}\,\blacksquare

    --

    A related problem.

    --

    As for \int_0^\infty {\frac{{\sin ^2 x}}<br />
{{x^2 }}\,dx} can be found here.
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  5. #5
    Math Engineering Student
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    Nice problem

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

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

    Substitute u=-\varphi,

    \int_{ - 1}^1 {\frac{2}<br />
{{\left( {u^2 + 3} \right)\left( {e^u + 1} \right)}}\,du} = \int_{ - 1}^1 {\frac{2}<br />
{{\left( {\varphi ^2 + 3} \right)\left( {e^{ - \varphi } + 1} \right)}}\,d\varphi ,} so

    \tau = \int_{ - 1}^1 {\frac{2}<br />
{{\left( {u^2 + 3} \right)\left( {e^u + 1} \right)}}\,du} = \int_{ - 1}^1 {\frac{2}<br />
{{\left( {u^2 + 3} \right)\left( {e^{ - u} + 1} \right)}}\,du} .

    And finally

    2\tau = \int_{ - 1}^1 {\frac{2}<br />
{{u^2 + 3}}\,du} = \frac{{2\pi }}<br />
{{3\sqrt 3 }}\,\therefore \,\tau = \frac{\pi }<br />
{{3\sqrt 3 }}.
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