I was trying to solve: ∫ 0 ∞ cos ⁡ ( x m ) x n d x

Question

I was trying to solve:                                    ∫                      0                                ∞                          cos        ⁡        (                  x          m                )                  x          n                                   d                x

cartonarepcpo4 · Accepted Answer

The integral is divergent if   m  =  0. Assume that   m  ≠  0. With the substitution       x    m    =  t, as you proposed, we get      ∫    0          +      ∞            cos    ⁡    (          x      m        )          x      n        d    x    =      1    m        ∫    0          +      ∞            cos    ⁡    (    t    )          t              (        n        +        1        )                  /                m        −        1              d    t    .Using the known Mellin transform formula for the cosine, we then find      1    m        ∫    0          +      ∞            cos    ⁡    (    t    )          t              (        n        +        1        )                  /                m        −        1              d    t    =      1    m    Γ        (                            n          +          1                m              )    cos  ⁡      (                            n          +          1                          2          m                    π        )  provided that   0  &amp;lt;            n      +      1        m    &amp;lt;  1

hyprkathknmk · Accepted Answer

You can also do it using the exponential integral function  ∫  cos  ⁡  (      x    m    )        x    n      d  x  =      1    m    ∫  cos  ⁡  (  t  )        t    k      d  t    with    k  =            n      −      m      +      1        m    I  (  t  )  =      1    m    ∫  cos  ⁡  (  t  )        t    k      d  t  =  −      1          2      m            t          k      +      1                  [            E          −      k        (  −  i  t  )  +      E          −      k        (  i  t  )            ]        I  (  0  )  =  −      i          2      m            (          e              +                  1          2                i        π        k              −          e              −                  1          2                i        π        k              )    Γ  (  k  +  1  )  =      1    m    sin  ⁡      (                  π        k            2        )    Γ  (  k  +  1  )Back to k  I  (  0  )  =  −      1    m    cos  ⁡      (                  n        +        1                    2        m              π    )      Γ      (                  n        +        1            m        )  At the upper bound, the convergence (to 0) requires   k  &amp;lt;  0 that is to say             n      +      1        m    &amp;lt;  1  if    0  &amp;lt;            n      +      1        m    &amp;lt;  1    then        ∫    0    ∞    cos  ⁡  (      x    m    )        x    n      d  x  =      1    m    cos  ⁡      (                  n        +        1                    2        m              π    )      Γ      (                  n        +        1            m        )

I was trying to solve: <mtable displaystyle="true"> <mlabeledtr> <mtd id="mjx-eqn-" />

Answered question

Answer & Explanation

New Questions in Integral Calculus