Is there a name for the integral $int_1^infty e^{-a(x+frac{b}{x})}dx$?












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It is known that the integral $int_0^infty e^{-a(x+frac{b}{x})}dx$ where $a$ and $b$ are two constants is BesselK-like function.



Is there a name for the integral $int_1^infty e^{-a(x+frac{b}{x})}dx$ ?










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  • $begingroup$
    Note: $int_1^infty f(x) dx = int_0^infty f(x) dx - int_0^1 f(x) dx$
    $endgroup$
    – DavidG
    Dec 28 '18 at 4:27






  • 4




    $begingroup$
    This integral corresponds to the incomplete Bessel function or "leaky aquifer function" $K_{-1}(a,ab)$ as defined by Harris. You can find details of the properties of this function in many papers by Harris, Fripiat and Jones (with a slightly different definition in the latter case).
    $endgroup$
    – Paul Enta
    Dec 28 '18 at 15:09










  • $begingroup$
    @PaulEnta - You should post your comment as a solution.
    $endgroup$
    – DavidG
    Dec 29 '18 at 3:51
















1












$begingroup$


It is known that the integral $int_0^infty e^{-a(x+frac{b}{x})}dx$ where $a$ and $b$ are two constants is BesselK-like function.



Is there a name for the integral $int_1^infty e^{-a(x+frac{b}{x})}dx$ ?










share|cite|improve this question









$endgroup$












  • $begingroup$
    Note: $int_1^infty f(x) dx = int_0^infty f(x) dx - int_0^1 f(x) dx$
    $endgroup$
    – DavidG
    Dec 28 '18 at 4:27






  • 4




    $begingroup$
    This integral corresponds to the incomplete Bessel function or "leaky aquifer function" $K_{-1}(a,ab)$ as defined by Harris. You can find details of the properties of this function in many papers by Harris, Fripiat and Jones (with a slightly different definition in the latter case).
    $endgroup$
    – Paul Enta
    Dec 28 '18 at 15:09










  • $begingroup$
    @PaulEnta - You should post your comment as a solution.
    $endgroup$
    – DavidG
    Dec 29 '18 at 3:51














1












1








1


1



$begingroup$


It is known that the integral $int_0^infty e^{-a(x+frac{b}{x})}dx$ where $a$ and $b$ are two constants is BesselK-like function.



Is there a name for the integral $int_1^infty e^{-a(x+frac{b}{x})}dx$ ?










share|cite|improve this question









$endgroup$




It is known that the integral $int_0^infty e^{-a(x+frac{b}{x})}dx$ where $a$ and $b$ are two constants is BesselK-like function.



Is there a name for the integral $int_1^infty e^{-a(x+frac{b}{x})}dx$ ?







integration definite-integrals bessel-functions






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asked Dec 27 '18 at 21:16









Anna NoieAnna Noie

733




733












  • $begingroup$
    Note: $int_1^infty f(x) dx = int_0^infty f(x) dx - int_0^1 f(x) dx$
    $endgroup$
    – DavidG
    Dec 28 '18 at 4:27






  • 4




    $begingroup$
    This integral corresponds to the incomplete Bessel function or "leaky aquifer function" $K_{-1}(a,ab)$ as defined by Harris. You can find details of the properties of this function in many papers by Harris, Fripiat and Jones (with a slightly different definition in the latter case).
    $endgroup$
    – Paul Enta
    Dec 28 '18 at 15:09










  • $begingroup$
    @PaulEnta - You should post your comment as a solution.
    $endgroup$
    – DavidG
    Dec 29 '18 at 3:51


















  • $begingroup$
    Note: $int_1^infty f(x) dx = int_0^infty f(x) dx - int_0^1 f(x) dx$
    $endgroup$
    – DavidG
    Dec 28 '18 at 4:27






  • 4




    $begingroup$
    This integral corresponds to the incomplete Bessel function or "leaky aquifer function" $K_{-1}(a,ab)$ as defined by Harris. You can find details of the properties of this function in many papers by Harris, Fripiat and Jones (with a slightly different definition in the latter case).
    $endgroup$
    – Paul Enta
    Dec 28 '18 at 15:09










  • $begingroup$
    @PaulEnta - You should post your comment as a solution.
    $endgroup$
    – DavidG
    Dec 29 '18 at 3:51
















$begingroup$
Note: $int_1^infty f(x) dx = int_0^infty f(x) dx - int_0^1 f(x) dx$
$endgroup$
– DavidG
Dec 28 '18 at 4:27




$begingroup$
Note: $int_1^infty f(x) dx = int_0^infty f(x) dx - int_0^1 f(x) dx$
$endgroup$
– DavidG
Dec 28 '18 at 4:27




4




4




$begingroup$
This integral corresponds to the incomplete Bessel function or "leaky aquifer function" $K_{-1}(a,ab)$ as defined by Harris. You can find details of the properties of this function in many papers by Harris, Fripiat and Jones (with a slightly different definition in the latter case).
$endgroup$
– Paul Enta
Dec 28 '18 at 15:09




$begingroup$
This integral corresponds to the incomplete Bessel function or "leaky aquifer function" $K_{-1}(a,ab)$ as defined by Harris. You can find details of the properties of this function in many papers by Harris, Fripiat and Jones (with a slightly different definition in the latter case).
$endgroup$
– Paul Enta
Dec 28 '18 at 15:09












$begingroup$
@PaulEnta - You should post your comment as a solution.
$endgroup$
– DavidG
Dec 29 '18 at 3:51




$begingroup$
@PaulEnta - You should post your comment as a solution.
$endgroup$
– DavidG
Dec 29 '18 at 3:51










1 Answer
1






active

oldest

votes


















1












$begingroup$

This integral corresponds to the incomplete Bessel function or leaky aquifer function $K_{-1}(a,ab)$ as defined by Harris:
begin{equation}
K_nu(x,y)=int_1^infty e^{-xt-frac{y}{t}},frac{dt}{t^{nu+1}}
end{equation}

Hydologist showed that ``water levels in pumped aquifer systems with finite transmissivity and leakage could be analyzed in terms of'' this integral.
You can find many properties of this function, as well as numerical methods to compute it in different regions in the $(a,b)$-plane, in papers by Harris and Fripiat. Jones uses a different (but related) definition:
begin{equation}
K_nu(z,w)=int_w^infty e^{-zcosh t}cosh nu t,dt
end{equation}

and offers a detailed analysis of its asymptotic expansion for complex values of the parameters.






share|cite|improve this answer









$endgroup$













  • $begingroup$
    You've saved my life Paul Enta :)
    $endgroup$
    – Anna Noie
    Dec 29 '18 at 10:25










  • $begingroup$
    My pleasure, enjoy!
    $endgroup$
    – Paul Enta
    Dec 29 '18 at 10:33











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1 Answer
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active

oldest

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1 Answer
1






active

oldest

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active

oldest

votes






active

oldest

votes









1












$begingroup$

This integral corresponds to the incomplete Bessel function or leaky aquifer function $K_{-1}(a,ab)$ as defined by Harris:
begin{equation}
K_nu(x,y)=int_1^infty e^{-xt-frac{y}{t}},frac{dt}{t^{nu+1}}
end{equation}

Hydologist showed that ``water levels in pumped aquifer systems with finite transmissivity and leakage could be analyzed in terms of'' this integral.
You can find many properties of this function, as well as numerical methods to compute it in different regions in the $(a,b)$-plane, in papers by Harris and Fripiat. Jones uses a different (but related) definition:
begin{equation}
K_nu(z,w)=int_w^infty e^{-zcosh t}cosh nu t,dt
end{equation}

and offers a detailed analysis of its asymptotic expansion for complex values of the parameters.






share|cite|improve this answer









$endgroup$













  • $begingroup$
    You've saved my life Paul Enta :)
    $endgroup$
    – Anna Noie
    Dec 29 '18 at 10:25










  • $begingroup$
    My pleasure, enjoy!
    $endgroup$
    – Paul Enta
    Dec 29 '18 at 10:33
















1












$begingroup$

This integral corresponds to the incomplete Bessel function or leaky aquifer function $K_{-1}(a,ab)$ as defined by Harris:
begin{equation}
K_nu(x,y)=int_1^infty e^{-xt-frac{y}{t}},frac{dt}{t^{nu+1}}
end{equation}

Hydologist showed that ``water levels in pumped aquifer systems with finite transmissivity and leakage could be analyzed in terms of'' this integral.
You can find many properties of this function, as well as numerical methods to compute it in different regions in the $(a,b)$-plane, in papers by Harris and Fripiat. Jones uses a different (but related) definition:
begin{equation}
K_nu(z,w)=int_w^infty e^{-zcosh t}cosh nu t,dt
end{equation}

and offers a detailed analysis of its asymptotic expansion for complex values of the parameters.






share|cite|improve this answer









$endgroup$













  • $begingroup$
    You've saved my life Paul Enta :)
    $endgroup$
    – Anna Noie
    Dec 29 '18 at 10:25










  • $begingroup$
    My pleasure, enjoy!
    $endgroup$
    – Paul Enta
    Dec 29 '18 at 10:33














1












1








1





$begingroup$

This integral corresponds to the incomplete Bessel function or leaky aquifer function $K_{-1}(a,ab)$ as defined by Harris:
begin{equation}
K_nu(x,y)=int_1^infty e^{-xt-frac{y}{t}},frac{dt}{t^{nu+1}}
end{equation}

Hydologist showed that ``water levels in pumped aquifer systems with finite transmissivity and leakage could be analyzed in terms of'' this integral.
You can find many properties of this function, as well as numerical methods to compute it in different regions in the $(a,b)$-plane, in papers by Harris and Fripiat. Jones uses a different (but related) definition:
begin{equation}
K_nu(z,w)=int_w^infty e^{-zcosh t}cosh nu t,dt
end{equation}

and offers a detailed analysis of its asymptotic expansion for complex values of the parameters.






share|cite|improve this answer









$endgroup$



This integral corresponds to the incomplete Bessel function or leaky aquifer function $K_{-1}(a,ab)$ as defined by Harris:
begin{equation}
K_nu(x,y)=int_1^infty e^{-xt-frac{y}{t}},frac{dt}{t^{nu+1}}
end{equation}

Hydologist showed that ``water levels in pumped aquifer systems with finite transmissivity and leakage could be analyzed in terms of'' this integral.
You can find many properties of this function, as well as numerical methods to compute it in different regions in the $(a,b)$-plane, in papers by Harris and Fripiat. Jones uses a different (but related) definition:
begin{equation}
K_nu(z,w)=int_w^infty e^{-zcosh t}cosh nu t,dt
end{equation}

and offers a detailed analysis of its asymptotic expansion for complex values of the parameters.







share|cite|improve this answer












share|cite|improve this answer



share|cite|improve this answer










answered Dec 29 '18 at 10:16









Paul EntaPaul Enta

5,16111334




5,16111334












  • $begingroup$
    You've saved my life Paul Enta :)
    $endgroup$
    – Anna Noie
    Dec 29 '18 at 10:25










  • $begingroup$
    My pleasure, enjoy!
    $endgroup$
    – Paul Enta
    Dec 29 '18 at 10:33


















  • $begingroup$
    You've saved my life Paul Enta :)
    $endgroup$
    – Anna Noie
    Dec 29 '18 at 10:25










  • $begingroup$
    My pleasure, enjoy!
    $endgroup$
    – Paul Enta
    Dec 29 '18 at 10:33
















$begingroup$
You've saved my life Paul Enta :)
$endgroup$
– Anna Noie
Dec 29 '18 at 10:25




$begingroup$
You've saved my life Paul Enta :)
$endgroup$
– Anna Noie
Dec 29 '18 at 10:25












$begingroup$
My pleasure, enjoy!
$endgroup$
– Paul Enta
Dec 29 '18 at 10:33




$begingroup$
My pleasure, enjoy!
$endgroup$
– Paul Enta
Dec 29 '18 at 10:33


















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