Solve ∫ 0 ∞ 1 ( x 2 + b 2 ) 4 d x

Question

Solve        ∫          0        ∞        1          (              x        2            +              b        2                    )        4                 d  x

Esteban Johnson · Accepted Answer

Integrate            (              t                  (          1          +                      t            2                                )                          n              −              1                                          )        ′    =            3      −      2      n              (      1      +              t        2                    )                  n          −          1                      +            2      n      −      2              (      1      +              t        2                    )                  n                    to establish       I    n    =      ∫    0    ∞        1          (      1      +              t        2                    )                  n                      d  t  =            2      n      −      3              2      n      −      2            I          n      −      1      . Then                              ∫          0          ∞                                      d            x                                (                          x              2                        +                          b              2                                      )              4                                      =                  1                      b            7                                    ∫          0                      ∞                                                d            t                                (                          t              2                        +            1                          )              4                                      =                  1                      b            7                                    5          6                          3          4                          1          2                          ∫          0          ∞                                      d            t                                1            +                          t              2                                      =                              5            π                                32                          b              7

Mohamed Mooney · Accepted Answer

This is the more bone-headed approach, but it works.
Letting

x = b t,

you get the integral

\begin{matrix} (1) & \frac{1}{b^{7}} \int_{0}^{\infty} \frac{1}{(t^{2} + 1)^{4}} d t \end{matrix}

Using Wolfram Alpha, I get the partial fractions:

\frac{1}{(t^{2} + 1)^{4}} = \frac{5 i}{32} (\frac{1}{t + i} - \frac{1}{t - i}) - \frac{5}{32} (\frac{1}{(t + i)^{2}} + \frac{1}{(t - i)^{2}}) - \frac{i}{8} (\frac{1}{(t + i)^{3}} - \frac{1}{(t - i)^{3}}) + \frac{1}{16} (\frac{1}{(t + i)^{4}} + \frac{1}{(t - i)^{4}})

The pairs cancel out in the integral from 0 to

\infty,

except for the first terms.
And

\frac{5 i}{32} (\frac{1}{t + i} - \frac{1}{t - i}) = \frac{5}{16} \frac{1}{t^{2} + 1}

So you get the result:

\frac{5}{16 b^{7}} \int_{0}^{\infty} \frac{d t}{t^{2} + 1} = \frac{5 π}{32 b^{7}}

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