Evaluating what looks like a Frullani integral












2












$begingroup$


Can the following integral be evaluated?



$$I=int_{0}^{infty} frac{e^{-ax^alpha}-e^{-bx^beta}}{x}dx$$



It looks to me as though it could be a Frullani integral, but I can't seem to find a good way to solve it (if there is one). I've even tried differentiating under the integral sign, but I couldn't seem to get anywhere with that either. If someone could be nice and either point me in the correct direction or just answer the question entirely, that would be very much appreciated. Thanks!






calculus







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$endgroup$












  • $begingroup$
    Is it meant to be $alpha x^alpha$ or is it intentionally two different constants?
    $endgroup$
    – Henry Lee
    Jan 21 at 0:17






  • 1




    $begingroup$
    Two different constants, sorry my code was a little bad so the integral looks tiny for some reason.
    $endgroup$
    – DerpyMcDerp
    Jan 21 at 0:31










  • $begingroup$
    I don’t believe is it of this form as we cannot define a function f(x) that fits both parts of the equation
    $endgroup$
    – Henry Lee
    Jan 21 at 0:48


















2












$begingroup$


Can the following integral be evaluated?



$$I=int_{0}^{infty} frac{e^{-ax^alpha}-e^{-bx^beta}}{x}dx$$



It looks to me as though it could be a Frullani integral, but I can't seem to find a good way to solve it (if there is one). I've even tried differentiating under the integral sign, but I couldn't seem to get anywhere with that either. If someone could be nice and either point me in the correct direction or just answer the question entirely, that would be very much appreciated. Thanks!






calculus







share|cite|improve this question











$endgroup$












  • $begingroup$
    Is it meant to be $alpha x^alpha$ or is it intentionally two different constants?
    $endgroup$
    – Henry Lee
    Jan 21 at 0:17






  • 1




    $begingroup$
    Two different constants, sorry my code was a little bad so the integral looks tiny for some reason.
    $endgroup$
    – DerpyMcDerp
    Jan 21 at 0:31










  • $begingroup$
    I don’t believe is it of this form as we cannot define a function f(x) that fits both parts of the equation
    $endgroup$
    – Henry Lee
    Jan 21 at 0:48
















2












2








2





$begingroup$


Can the following integral be evaluated?



$$I=int_{0}^{infty} frac{e^{-ax^alpha}-e^{-bx^beta}}{x}dx$$



It looks to me as though it could be a Frullani integral, but I can't seem to find a good way to solve it (if there is one). I've even tried differentiating under the integral sign, but I couldn't seem to get anywhere with that either. If someone could be nice and either point me in the correct direction or just answer the question entirely, that would be very much appreciated. Thanks!






calculus







share|cite|improve this question











$endgroup$




Can the following integral be evaluated?



$$I=int_{0}^{infty} frac{e^{-ax^alpha}-e^{-bx^beta}}{x}dx$$



It looks to me as though it could be a Frullani integral, but I can't seem to find a good way to solve it (if there is one). I've even tried differentiating under the integral sign, but I couldn't seem to get anywhere with that either. If someone could be nice and either point me in the correct direction or just answer the question entirely, that would be very much appreciated. Thanks!






calculus




calculus






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Jan 21 at 0:33









Henry Lee

2,042219




2,042219










asked Jan 21 at 0:07









DerpyMcDerpDerpyMcDerp

905




905












  • $begingroup$
    Is it meant to be $alpha x^alpha$ or is it intentionally two different constants?
    $endgroup$
    – Henry Lee
    Jan 21 at 0:17






  • 1




    $begingroup$
    Two different constants, sorry my code was a little bad so the integral looks tiny for some reason.
    $endgroup$
    – DerpyMcDerp
    Jan 21 at 0:31










  • $begingroup$
    I don’t believe is it of this form as we cannot define a function f(x) that fits both parts of the equation
    $endgroup$
    – Henry Lee
    Jan 21 at 0:48




















  • $begingroup$
    Is it meant to be $alpha x^alpha$ or is it intentionally two different constants?
    $endgroup$
    – Henry Lee
    Jan 21 at 0:17






  • 1




    $begingroup$
    Two different constants, sorry my code was a little bad so the integral looks tiny for some reason.
    $endgroup$
    – DerpyMcDerp
    Jan 21 at 0:31










  • $begingroup$
    I don’t believe is it of this form as we cannot define a function f(x) that fits both parts of the equation
    $endgroup$
    – Henry Lee
    Jan 21 at 0:48


















$begingroup$
Is it meant to be $alpha x^alpha$ or is it intentionally two different constants?
$endgroup$
– Henry Lee
Jan 21 at 0:17




$begingroup$
Is it meant to be $alpha x^alpha$ or is it intentionally two different constants?
$endgroup$
– Henry Lee
Jan 21 at 0:17




1




1




$begingroup$
Two different constants, sorry my code was a little bad so the integral looks tiny for some reason.
$endgroup$
– DerpyMcDerp
Jan 21 at 0:31




$begingroup$
Two different constants, sorry my code was a little bad so the integral looks tiny for some reason.
$endgroup$
– DerpyMcDerp
Jan 21 at 0:31












$begingroup$
I don’t believe is it of this form as we cannot define a function f(x) that fits both parts of the equation
$endgroup$
– Henry Lee
Jan 21 at 0:48






$begingroup$
I don’t believe is it of this form as we cannot define a function f(x) that fits both parts of the equation
$endgroup$
– Henry Lee
Jan 21 at 0:48












1 Answer
1






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oldest

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4












$begingroup$

Let
$$I = int_0^infty frac{e^{-ax^alpha} - e^{-b x^beta}}{x} , dx.$$
I will assume $a,b, alpha, beta > 0$.



Integrating by parts we have
$$I = a alpha int_0^infty ln x , x^{alpha - 1} e^{-a x^alpha} , dx - b beta int_0^infty ln x, x^{beta - 1} e^{-b x^beta} , dx = I_1 - I_2.$$



In $I_1$ setting $u = ax^alpha$, gives $dx = (u/a)^{frac{1}{alpha} - 1} du/(aalpha)$. The limits of integral are unchanged since $alpha > 0$ (see below if $alpha, beta < 0$). Thus
begin{align}
I_1 &= frac{1}{alpha} int_0^infty ln left (frac{u}{a} right ) e^{-u} , du\
&= frac{1}{alpha} int_0^infty ln u e^{-u} , du - frac{ln a}{alpha} int_0^infty e^{-u} , du\
&= frac{1}{alpha} (-gamma - ln a).
end{align}
Here the integral representation for the Euler–Mascheroni constant of
$$gamma = - int_0^infty ln x e^{-x} , dx,$$
has been used.



Similarly
$I_2 = frac{1}{beta} (-gamma - ln b).$
Thus
$$int_0^infty frac{e^{-ax^alpha} - e^{-b x^beta}}{x} , dx = frac{1}{alpha} (-gamma - ln a) - frac{1}{beta} (-gamma - ln b).$$



Note the result can be extended to all $alpha, beta in mathbb{R}$ provided the product $alpha beta$ is positive (needed if we are to have convergence in the integral). In this case the limits of integration become reversed in the substitution made in $I_1$. The final result is therefore
$$int_0^infty frac{e^{-ax^alpha} - e^{-b x^beta}}{x} , dx = frac{1}{|alpha|} (-gamma - ln a) - frac{1}{|beta|} (-gamma - ln b), qquad a,b > 0, alpha beta > 0.$$






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    4












    $begingroup$

    Let
    $$I = int_0^infty frac{e^{-ax^alpha} - e^{-b x^beta}}{x} , dx.$$
    I will assume $a,b, alpha, beta > 0$.



    Integrating by parts we have
    $$I = a alpha int_0^infty ln x , x^{alpha - 1} e^{-a x^alpha} , dx - b beta int_0^infty ln x, x^{beta - 1} e^{-b x^beta} , dx = I_1 - I_2.$$



    In $I_1$ setting $u = ax^alpha$, gives $dx = (u/a)^{frac{1}{alpha} - 1} du/(aalpha)$. The limits of integral are unchanged since $alpha > 0$ (see below if $alpha, beta < 0$). Thus
    begin{align}
    I_1 &= frac{1}{alpha} int_0^infty ln left (frac{u}{a} right ) e^{-u} , du\
    &= frac{1}{alpha} int_0^infty ln u e^{-u} , du - frac{ln a}{alpha} int_0^infty e^{-u} , du\
    &= frac{1}{alpha} (-gamma - ln a).
    end{align}
    Here the integral representation for the Euler–Mascheroni constant of
    $$gamma = - int_0^infty ln x e^{-x} , dx,$$
    has been used.



    Similarly
    $I_2 = frac{1}{beta} (-gamma - ln b).$
    Thus
    $$int_0^infty frac{e^{-ax^alpha} - e^{-b x^beta}}{x} , dx = frac{1}{alpha} (-gamma - ln a) - frac{1}{beta} (-gamma - ln b).$$



    Note the result can be extended to all $alpha, beta in mathbb{R}$ provided the product $alpha beta$ is positive (needed if we are to have convergence in the integral). In this case the limits of integration become reversed in the substitution made in $I_1$. The final result is therefore
    $$int_0^infty frac{e^{-ax^alpha} - e^{-b x^beta}}{x} , dx = frac{1}{|alpha|} (-gamma - ln a) - frac{1}{|beta|} (-gamma - ln b), qquad a,b > 0, alpha beta > 0.$$






    share|cite|improve this answer











    $endgroup$


















      4












      $begingroup$

      Let
      $$I = int_0^infty frac{e^{-ax^alpha} - e^{-b x^beta}}{x} , dx.$$
      I will assume $a,b, alpha, beta > 0$.



      Integrating by parts we have
      $$I = a alpha int_0^infty ln x , x^{alpha - 1} e^{-a x^alpha} , dx - b beta int_0^infty ln x, x^{beta - 1} e^{-b x^beta} , dx = I_1 - I_2.$$



      In $I_1$ setting $u = ax^alpha$, gives $dx = (u/a)^{frac{1}{alpha} - 1} du/(aalpha)$. The limits of integral are unchanged since $alpha > 0$ (see below if $alpha, beta < 0$). Thus
      begin{align}
      I_1 &= frac{1}{alpha} int_0^infty ln left (frac{u}{a} right ) e^{-u} , du\
      &= frac{1}{alpha} int_0^infty ln u e^{-u} , du - frac{ln a}{alpha} int_0^infty e^{-u} , du\
      &= frac{1}{alpha} (-gamma - ln a).
      end{align}
      Here the integral representation for the Euler–Mascheroni constant of
      $$gamma = - int_0^infty ln x e^{-x} , dx,$$
      has been used.



      Similarly
      $I_2 = frac{1}{beta} (-gamma - ln b).$
      Thus
      $$int_0^infty frac{e^{-ax^alpha} - e^{-b x^beta}}{x} , dx = frac{1}{alpha} (-gamma - ln a) - frac{1}{beta} (-gamma - ln b).$$



      Note the result can be extended to all $alpha, beta in mathbb{R}$ provided the product $alpha beta$ is positive (needed if we are to have convergence in the integral). In this case the limits of integration become reversed in the substitution made in $I_1$. The final result is therefore
      $$int_0^infty frac{e^{-ax^alpha} - e^{-b x^beta}}{x} , dx = frac{1}{|alpha|} (-gamma - ln a) - frac{1}{|beta|} (-gamma - ln b), qquad a,b > 0, alpha beta > 0.$$






      share|cite|improve this answer











      $endgroup$
















        4












        4








        4





        $begingroup$

        Let
        $$I = int_0^infty frac{e^{-ax^alpha} - e^{-b x^beta}}{x} , dx.$$
        I will assume $a,b, alpha, beta > 0$.



        Integrating by parts we have
        $$I = a alpha int_0^infty ln x , x^{alpha - 1} e^{-a x^alpha} , dx - b beta int_0^infty ln x, x^{beta - 1} e^{-b x^beta} , dx = I_1 - I_2.$$



        In $I_1$ setting $u = ax^alpha$, gives $dx = (u/a)^{frac{1}{alpha} - 1} du/(aalpha)$. The limits of integral are unchanged since $alpha > 0$ (see below if $alpha, beta < 0$). Thus
        begin{align}
        I_1 &= frac{1}{alpha} int_0^infty ln left (frac{u}{a} right ) e^{-u} , du\
        &= frac{1}{alpha} int_0^infty ln u e^{-u} , du - frac{ln a}{alpha} int_0^infty e^{-u} , du\
        &= frac{1}{alpha} (-gamma - ln a).
        end{align}
        Here the integral representation for the Euler–Mascheroni constant of
        $$gamma = - int_0^infty ln x e^{-x} , dx,$$
        has been used.



        Similarly
        $I_2 = frac{1}{beta} (-gamma - ln b).$
        Thus
        $$int_0^infty frac{e^{-ax^alpha} - e^{-b x^beta}}{x} , dx = frac{1}{alpha} (-gamma - ln a) - frac{1}{beta} (-gamma - ln b).$$



        Note the result can be extended to all $alpha, beta in mathbb{R}$ provided the product $alpha beta$ is positive (needed if we are to have convergence in the integral). In this case the limits of integration become reversed in the substitution made in $I_1$. The final result is therefore
        $$int_0^infty frac{e^{-ax^alpha} - e^{-b x^beta}}{x} , dx = frac{1}{|alpha|} (-gamma - ln a) - frac{1}{|beta|} (-gamma - ln b), qquad a,b > 0, alpha beta > 0.$$






        share|cite|improve this answer











        $endgroup$



        Let
        $$I = int_0^infty frac{e^{-ax^alpha} - e^{-b x^beta}}{x} , dx.$$
        I will assume $a,b, alpha, beta > 0$.



        Integrating by parts we have
        $$I = a alpha int_0^infty ln x , x^{alpha - 1} e^{-a x^alpha} , dx - b beta int_0^infty ln x, x^{beta - 1} e^{-b x^beta} , dx = I_1 - I_2.$$



        In $I_1$ setting $u = ax^alpha$, gives $dx = (u/a)^{frac{1}{alpha} - 1} du/(aalpha)$. The limits of integral are unchanged since $alpha > 0$ (see below if $alpha, beta < 0$). Thus
        begin{align}
        I_1 &= frac{1}{alpha} int_0^infty ln left (frac{u}{a} right ) e^{-u} , du\
        &= frac{1}{alpha} int_0^infty ln u e^{-u} , du - frac{ln a}{alpha} int_0^infty e^{-u} , du\
        &= frac{1}{alpha} (-gamma - ln a).
        end{align}
        Here the integral representation for the Euler–Mascheroni constant of
        $$gamma = - int_0^infty ln x e^{-x} , dx,$$
        has been used.



        Similarly
        $I_2 = frac{1}{beta} (-gamma - ln b).$
        Thus
        $$int_0^infty frac{e^{-ax^alpha} - e^{-b x^beta}}{x} , dx = frac{1}{alpha} (-gamma - ln a) - frac{1}{beta} (-gamma - ln b).$$



        Note the result can be extended to all $alpha, beta in mathbb{R}$ provided the product $alpha beta$ is positive (needed if we are to have convergence in the integral). In this case the limits of integration become reversed in the substitution made in $I_1$. The final result is therefore
        $$int_0^infty frac{e^{-ax^alpha} - e^{-b x^beta}}{x} , dx = frac{1}{|alpha|} (-gamma - ln a) - frac{1}{|beta|} (-gamma - ln b), qquad a,b > 0, alpha beta > 0.$$







        share|cite|improve this answer














        share|cite|improve this answer



        share|cite|improve this answer








        edited Jan 21 at 5:19

























        answered Jan 21 at 1:59









        omegadotomegadot

        6,0222828




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