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Math Help - Strugling with an integral

  1. #1
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    Strugling with an integral

    \int^a_\infty \frac{dz}{(1+z)^2(\Omega_0z+1)^{\frac12}}

    It is quite straightforward when \Omega=1 but I am strugling when \Omega>1.

    Can I split this into partial fractions somehow? Or perhaps use integration by parts twice? Neither options seem to work for me.
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  2. #2
    Senior Member DeMath's Avatar
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    Quote Originally Posted by Kiwi_Dave View Post
    \int^a_\infty \frac{dz}{(1+z)^2(\Omega_0z+1)^{\frac12}}

    It is quite straightforward when \Omega=1 but I am strugling when \Omega>1.

    Can I split this into partial fractions somehow? Or perhaps use integration by parts twice? Neither options seem to work for me.
    \int {\frac{{dz}}{{{{\left( {1 + z} \right)}^2}\sqrt {az + 1} }}}  = \left\{ \begin{gathered}az + 1 = {t^2}, \hfill \\z = \frac{{{t^2} - 1}}<br />
{a}, \hfill \\dz = \frac{{2t}}{a}dt \hfill \\ \end{gathered}  \right\} = \frac{2}{a}\int {\frac{{dt}}{{{{\left( {1 + \frac{{{t^2} - 1}}{a}} \right)}^2}}}}  =

    = 2a\int {\frac{{dt}}<br />
{{{{\left( {a - 1 + {t^2}} \right)}^2}}}}  = \frac{{2a}}<br />
{{{{\left( {a - 1} \right)}^2}}}\int {\frac{{dt}}<br />
{{{{\left( {1 + \frac{{{t^2}}}<br />
{{a - 1}}} \right)}^2}}}}  = \left\{ \begin{gathered}<br />
  t = \sqrt {a - 1} \tan u, \hfill \\<br />
  dt = \frac{{\sqrt {a - 1} }}<br />
{{{{\cos }^2}u}}du \hfill \\ <br />
\end{gathered}  \right\} =

    = \frac{{2a\sqrt {a - 1} }}<br />
{{{{\left( {a - 1} \right)}^2}}}\int {\frac{{du}}<br />
{{{{\left( {1 + {{\tan }^2}u} \right)}^2}{{\cos }^2}u}}}  = \frac{{2a}}<br />
{{\sqrt {{{\left( {a - 1} \right)}^3}} }}\int {{{\cos }^2}udu}  =

    = \frac{a}{{\sqrt {{{\left( {a - 1} \right)}^3}} }}\int {\left( {1 + \cos 2u} \right)du}  = \frac{a}{{\sqrt {{{\left( {a - 1} \right)}^3}} }}\left( {u + \frac{1}{2}\sin 2u} \right) + C =

    = \frac{a}<br />
{{\sqrt {{{\left( {a - 1} \right)}^3}} }}\left[ {\arctan \frac{t}<br />
{{\sqrt {a - 1} }} + \frac{1}<br />
{2}\sin \left( {2\arctan \frac{t}<br />
{{\sqrt {a - 1} }}} \right)} \right] + C =

    = \frac{a}{{\sqrt {{{\left( {a - 1} \right)}^3}} }}\left[ {\arctan \sqrt {\frac{{az + 1}}{{a - 1}}}  + \frac{1}{2}\sin \left( {2\arctan \sqrt {\frac{{az + 1}}{{a - 1}}} } \right)} \right] + C.

    Simplification

    \frac{1}{2}\sin \left( {2\arctan \sqrt {\frac{{az + 1}}{{a - 1}}} } \right) = \sin \left( {\arctan \sqrt {\frac{{az + 1}}{{a - 1}}} } \right)\cos \left( {\arctan \sqrt {\frac{{az + 1}}{{a - 1}}} } \right) =

    = \frac{{\sqrt {\frac{{az + 1}}{{a - 1}}} }}{{\sqrt {1 + \frac{{az + 1}}{{a - 1}}} }} \cdot \frac{1}{{\sqrt {1 + \frac{{az + 1}}{{a - 1}}} }} = \frac{{\sqrt {\frac{{az + 1}}{{a - 1}}} }}{{1 + \frac{{az + 1}}{{a - 1}}}} = \frac{{\sqrt {az + 1} \sqrt {a - 1} }}{{a\left( {z + 1} \right)}}.

    Finally

    \int {\frac{{dz}}{{{{\left( {1 + z} \right)}^2}\sqrt {az + 1} }}}  = \frac{a}{{\sqrt {{{\left( {a - 1} \right)}^3}} }}\arctan \sqrt {\frac{{az + 1}}{{a - 1}}}  + \frac{{\sqrt {az + 1} }}{{\left( {z + 1} \right)\left( {a - 1} \right)}} + C.
    a \in (1; \, + \infty).
    Last edited by DeMath; November 7th 2009 at 01:13 PM.
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