Strugling with an integral

**Kiwi_Dave** · Oct 5th 2009, 12:57 AM

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

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.

**DeMath** · Oct 5th 2009, 12:10 PM

Originally Posted by Kiwi_Dave

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

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.

$\int {\frac{{dz}}{{{{\left( {1 + z} \right)}^2}\sqrt {az + 1} }}} = \left\{ \begin{gathered}az + 1 = {t^2}, \hfill \\z = \frac{{{t^2} - 1}} {a}, \hfill \\dz = \frac{{2t}}{a}dt \hfill \\ \end{gathered} \right\} = \frac{2}{a}\int {\frac{{dt}}{{{{\left( {1 + \frac{{{t^2} - 1}}{a}} \right)}^2}}}} =$

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

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

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

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

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

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

$\int {\frac{{dz}}{{{{\left( {1 + z} \right)}^2}\sqrt {az + 1} }}} = \frac{a}{{\sqrt {{{\left( {a - 1} \right)}^3}} }}\arctan \sqrt {\frac{{az + 1}}{{a - 1}}} + \frac{{\sqrt {az + 1} }}{{\left( {z + 1} \right)\left( {a - 1} \right)}} + C.$
$a \in (1; \, + \infty).$

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