I have the following expression: ∫ 0 L T x 2 2 × 10 − 4 e − 2 × 10 − 4 x d x + ∫ L T ∞ L T 2 2 × 10 − 4 e − 2 × 10 − 4 x d x = 7065080

Question

I have the following expression:      ∫          0              L      T            x    2    2  ×      10          −      4            e          −      2      ×              10                  −          4                    x        d  x  +      ∫          L      T              ∞        L      T    2    2  ×      10          −      4            e          −      2      ×              10                  −          4                    x        d  x  =  7065080

Sawyer Day · Accepted Answer

I wonder if there is or not a mistake (the       L    2   must disappear).      I    1    =      1    5000        ∫    0    L        x    2          e          −      x              /            5000          d  x  =  5  ×      10    7    −      e          −      L              /            5000        (  5  ×      10    7    +      10    4    L  +      L    2    )      I    2    =            L      2        5000        ∫    L    ∞        e          −      x              /            5000          d  x  =      e          −      L              /            5000            L    2                      (1)                    f        (        L        )        =                  I          1                +                  I          2                −        7065080        =        42934920        −                  10          4                                  e                      −            L                          /                        5000                          (        L        +        5000        )        =        0            The only explicit solution is given in terms of Lambert function  L  =  −  5000      (    1    +          W              −        1                    (      −              1073373                  1250000          e                    )        )    =  3289.36  ⋯What is amuzing is that expanding   (  1  ) as a Taylor series around   L  =  0  f  (  L  )  =  −  7065080  +      L    2    −            L      3        7500    +            L      4        100000000    +  O      (          L      5        )  and using series reversion  L  =  t  +            t      2        15000    +            11              t        3              1800000000    +  O      (          t      4        )      where    t  =      f    (    L    )    +    7065080  Since we want   f  (  L  )  =  0, then the approximation  L  ∼      176627    375    +      46942897    2250000              176627      10        =  3243.79  ⋯

I have the following expression: <msubsup> ∫ <mrow class="MJX-TeXAtom-O

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