A definite integral involve Logarithmic FunctionsHere is the integral body: ∫ 0 m x a ln ⁡ ( x + b ) d x , a &gt; − 3 2 , b &gt; 0

Question

A definite integral involve Logarithmic FunctionsHere is the integral body:      ∫    0    m                      x        a              ln    ⁡          (              x        +        b            )        d    x    ,    a    &amp;gt;    −          3      2        ,    b    &amp;gt;    0

RamPatWeese2w · Accepted Answer

As mentioned in the comments, integration by parts is a good place to start. The   a  =  −  1 case has to be handled separately (and it&#039;s not difficult to show the integral diverges in this case), but otherwise,  I  (  m  ,  a  ,  b  )  =      ∫          0              m            x    a    log  ⁡            (      x      +      b      )        d  x  =      1          a      +      1            x          a      +      1        log  ⁡            (      x      +      b      )                          |                    0              m        −      1          a      +      1            ∫          0              m                  x              a        +        1                    x      +      b        d  x    =                    m                  a          +          1                    log      ⁡                        (          m          +          b          )                            a      +      1        −      1          a      +      1            ∫          0              m                  x              a        +        1                    x      +      b        d  x  .Since by assumption   b  &amp;gt;  0, it&#039;s a good candidate for a rescaling argument. Substituting   x  =  b  u into the integral that resulted from IBPs abov:         d  x  =  b    d  u,   u  =      x    b  , and so      ∫          0              m                  x              a        +        1                    x      +      b        d  x  =      ∫          0                      m        b                        (      b      u              )                  a          +          1                            b      u      +      b        (  b    d  u  )    =      b          a      +      1            ∫          0                      m        b                        u              a        +        1                    u      +      1          d  u  .Simplifications past this this point will depend a great deal upon the individual&#039;s opinion about what constitutes &#039;simpler&#039;.

ndevunidt · Accepted Answer

∫      0      m                      x                  n          −          1                            x        +        a              d    x    =                  m        n                    a        ⋅        n              ⋅                                             2                            F                                  1                                    (              1    ,    n         ;         n    +    1         ;    −          m      a                      )              ,   where                                 2                  F                          1             is the Gaussian hypergeometric function. For large values of m, we have             ∫      0      ∞                      x        n                    x        +        a              d    x    =    −          a      n        ⋅    π    ⋅    csc    ⁡    (    n    π    )

A definite integral involve Logarithmic Functions int_0^m ((x^a)ln((x+b))dx,a>-(3)/(2),b > 0)

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