Attain a two-term approximation for the following integral as m goes to 1 from below: I = ∫ 0 π / 2 d θ 1 − ( m 2 ) ⋅ sin ⁡ ( θ ) 2 .

Question

Attain a two-term approximation for the following integral as   m goes to   1 from below:  I  =      ∫          0              π              /            2                          d            θ              1      −      (              m        2            )      ⋅      sin      ⁡      (      θ              )        2              .

Mekjulleymg · Accepted Answer

This is just the complete elliptic intagral of the first kind, thus  I  =  K  (  m  )  =  −      1          2      π            ∑          n      =      0        ∞                                Γ          2                                  (                      n            +                          1              2                                )                                      Γ          2                (        n        +        1        )              (    1    −          m      2              )      n        (    log    ⁡    (    1    −          m      2        )    +          a      n        )  as   m  →  1  −  0, where       a    0    =  2  log  ⁡  2 and      a          n      +      1        =      a    n    +      4          (      2      n      +      1      )      (      2      n      +      2      )          (  n  ≥  0  )  .

Attain a two-term approximation for the following integral as m goes to 1 from below:

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