Exponential integral times exponential












1












$begingroup$


Let us define the exponential integral $E_{1}(x) = int_{x}^{infty} frac{e^{-t}}{t} mathrm{d}t$, with $x$ positive real.



Is there any way to simplify the following product?



$$e^{x} E_{1}(x)$$



I've tried to rewrite the exponential integral using the incomplete gamma function, i.e., $E_{1}(x) = Gamma(0,x)$ but it does not simplify anything. I've also considered the "continued fraction expansion" in https://en.wikipedia.org/wiki/Exponential_integral#Approximations: if I multiply the exponential integral by the exponential, we get rid of the exponential in the numerator of the "continued fraction expansion", but the expression remains intractable.






real-analysis integration exponential-function







share|cite|improve this question









$endgroup$












  • $begingroup$
    What is the context of this problem?
    $endgroup$
    – D.R.
    Dec 12 '17 at 15:00










  • $begingroup$
    @D.R. Shannon capacity of MIMO communication channels.
    $endgroup$
    – TheDon
    Dec 12 '17 at 15:09








  • 1




    $begingroup$
    Well, according to the Wiki page you have $e^x E_1(x) = U(1,1,x)$ with Tricomi's confluent hypergeometric function, see also dlmf.nist.gov/6.11, and there is an asyptotic expression
    $endgroup$
    – gammatester
    Dec 12 '17 at 15:36










  • $begingroup$
    @gammatester Thanks. It really looks like there is no tractable expression, i.e., one that does not involved "masked integrals"...
    $endgroup$
    – TheDon
    Dec 12 '17 at 15:45






  • 1




    $begingroup$
    Define "simplify". The only way this would simplify is if $E_1(x)$ could simplify, and as far as I'm concerned, it doesn't, being a transcendental special function.
    $endgroup$
    – Simply Beautiful Art
    Dec 14 '17 at 1:59


















1












$begingroup$


Let us define the exponential integral $E_{1}(x) = int_{x}^{infty} frac{e^{-t}}{t} mathrm{d}t$, with $x$ positive real.



Is there any way to simplify the following product?



$$e^{x} E_{1}(x)$$



I've tried to rewrite the exponential integral using the incomplete gamma function, i.e., $E_{1}(x) = Gamma(0,x)$ but it does not simplify anything. I've also considered the "continued fraction expansion" in https://en.wikipedia.org/wiki/Exponential_integral#Approximations: if I multiply the exponential integral by the exponential, we get rid of the exponential in the numerator of the "continued fraction expansion", but the expression remains intractable.






real-analysis integration exponential-function







share|cite|improve this question









$endgroup$












  • $begingroup$
    What is the context of this problem?
    $endgroup$
    – D.R.
    Dec 12 '17 at 15:00










  • $begingroup$
    @D.R. Shannon capacity of MIMO communication channels.
    $endgroup$
    – TheDon
    Dec 12 '17 at 15:09








  • 1




    $begingroup$
    Well, according to the Wiki page you have $e^x E_1(x) = U(1,1,x)$ with Tricomi's confluent hypergeometric function, see also dlmf.nist.gov/6.11, and there is an asyptotic expression
    $endgroup$
    – gammatester
    Dec 12 '17 at 15:36










  • $begingroup$
    @gammatester Thanks. It really looks like there is no tractable expression, i.e., one that does not involved "masked integrals"...
    $endgroup$
    – TheDon
    Dec 12 '17 at 15:45






  • 1




    $begingroup$
    Define "simplify". The only way this would simplify is if $E_1(x)$ could simplify, and as far as I'm concerned, it doesn't, being a transcendental special function.
    $endgroup$
    – Simply Beautiful Art
    Dec 14 '17 at 1:59
















1












1








1





$begingroup$


Let us define the exponential integral $E_{1}(x) = int_{x}^{infty} frac{e^{-t}}{t} mathrm{d}t$, with $x$ positive real.



Is there any way to simplify the following product?



$$e^{x} E_{1}(x)$$



I've tried to rewrite the exponential integral using the incomplete gamma function, i.e., $E_{1}(x) = Gamma(0,x)$ but it does not simplify anything. I've also considered the "continued fraction expansion" in https://en.wikipedia.org/wiki/Exponential_integral#Approximations: if I multiply the exponential integral by the exponential, we get rid of the exponential in the numerator of the "continued fraction expansion", but the expression remains intractable.






real-analysis integration exponential-function







share|cite|improve this question









$endgroup$




Let us define the exponential integral $E_{1}(x) = int_{x}^{infty} frac{e^{-t}}{t} mathrm{d}t$, with $x$ positive real.



Is there any way to simplify the following product?



$$e^{x} E_{1}(x)$$



I've tried to rewrite the exponential integral using the incomplete gamma function, i.e., $E_{1}(x) = Gamma(0,x)$ but it does not simplify anything. I've also considered the "continued fraction expansion" in https://en.wikipedia.org/wiki/Exponential_integral#Approximations: if I multiply the exponential integral by the exponential, we get rid of the exponential in the numerator of the "continued fraction expansion", but the expression remains intractable.






real-analysis integration exponential-function




real-analysis integration exponential-function






share|cite|improve this question













share|cite|improve this question











share|cite|improve this question




share|cite|improve this question










asked Dec 12 '17 at 14:53









TheDonTheDon

317313




317313












  • $begingroup$
    What is the context of this problem?
    $endgroup$
    – D.R.
    Dec 12 '17 at 15:00










  • $begingroup$
    @D.R. Shannon capacity of MIMO communication channels.
    $endgroup$
    – TheDon
    Dec 12 '17 at 15:09








  • 1




    $begingroup$
    Well, according to the Wiki page you have $e^x E_1(x) = U(1,1,x)$ with Tricomi's confluent hypergeometric function, see also dlmf.nist.gov/6.11, and there is an asyptotic expression
    $endgroup$
    – gammatester
    Dec 12 '17 at 15:36










  • $begingroup$
    @gammatester Thanks. It really looks like there is no tractable expression, i.e., one that does not involved "masked integrals"...
    $endgroup$
    – TheDon
    Dec 12 '17 at 15:45






  • 1




    $begingroup$
    Define "simplify". The only way this would simplify is if $E_1(x)$ could simplify, and as far as I'm concerned, it doesn't, being a transcendental special function.
    $endgroup$
    – Simply Beautiful Art
    Dec 14 '17 at 1:59




















  • $begingroup$
    What is the context of this problem?
    $endgroup$
    – D.R.
    Dec 12 '17 at 15:00










  • $begingroup$
    @D.R. Shannon capacity of MIMO communication channels.
    $endgroup$
    – TheDon
    Dec 12 '17 at 15:09








  • 1




    $begingroup$
    Well, according to the Wiki page you have $e^x E_1(x) = U(1,1,x)$ with Tricomi's confluent hypergeometric function, see also dlmf.nist.gov/6.11, and there is an asyptotic expression
    $endgroup$
    – gammatester
    Dec 12 '17 at 15:36










  • $begingroup$
    @gammatester Thanks. It really looks like there is no tractable expression, i.e., one that does not involved "masked integrals"...
    $endgroup$
    – TheDon
    Dec 12 '17 at 15:45






  • 1




    $begingroup$
    Define "simplify". The only way this would simplify is if $E_1(x)$ could simplify, and as far as I'm concerned, it doesn't, being a transcendental special function.
    $endgroup$
    – Simply Beautiful Art
    Dec 14 '17 at 1:59


















$begingroup$
What is the context of this problem?
$endgroup$
– D.R.
Dec 12 '17 at 15:00




$begingroup$
What is the context of this problem?
$endgroup$
– D.R.
Dec 12 '17 at 15:00












$begingroup$
@D.R. Shannon capacity of MIMO communication channels.
$endgroup$
– TheDon
Dec 12 '17 at 15:09






$begingroup$
@D.R. Shannon capacity of MIMO communication channels.
$endgroup$
– TheDon
Dec 12 '17 at 15:09






1




1




$begingroup$
Well, according to the Wiki page you have $e^x E_1(x) = U(1,1,x)$ with Tricomi's confluent hypergeometric function, see also dlmf.nist.gov/6.11, and there is an asyptotic expression
$endgroup$
– gammatester
Dec 12 '17 at 15:36




$begingroup$
Well, according to the Wiki page you have $e^x E_1(x) = U(1,1,x)$ with Tricomi's confluent hypergeometric function, see also dlmf.nist.gov/6.11, and there is an asyptotic expression
$endgroup$
– gammatester
Dec 12 '17 at 15:36












$begingroup$
@gammatester Thanks. It really looks like there is no tractable expression, i.e., one that does not involved "masked integrals"...
$endgroup$
– TheDon
Dec 12 '17 at 15:45




$begingroup$
@gammatester Thanks. It really looks like there is no tractable expression, i.e., one that does not involved "masked integrals"...
$endgroup$
– TheDon
Dec 12 '17 at 15:45




1




1




$begingroup$
Define "simplify". The only way this would simplify is if $E_1(x)$ could simplify, and as far as I'm concerned, it doesn't, being a transcendental special function.
$endgroup$
– Simply Beautiful Art
Dec 14 '17 at 1:59






$begingroup$
Define "simplify". The only way this would simplify is if $E_1(x)$ could simplify, and as far as I'm concerned, it doesn't, being a transcendental special function.
$endgroup$
– Simply Beautiful Art
Dec 14 '17 at 1:59












2 Answers
2






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$begingroup$

If there was a known simplification, we would not be using the function $E_1$, but instead that simplified function times $e^{-x}$.






share|cite|improve this answer









$endgroup$





















    -1












    $begingroup$

    The only operation in the way to simplify this is integrating by parts the $E_1$:



    $$E_1left( xright) = int_x^infty dleft(e^{-t}lntright) + int_x^infty e^{-t}lnt dt = -e^{-x}lnx + int_x^infty e^{-t}lnt dt$$



    Then:



    $$e^xE_1left( x right) = -lnx + e^xint_x^infty e^{-t}lnt dt$$






    share|cite|improve this answer











    $endgroup$













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      2 Answers
      2






      active

      oldest

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      2 Answers
      2






      active

      oldest

      votes









      active

      oldest

      votes






      active

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      votes









      0












      $begingroup$

      If there was a known simplification, we would not be using the function $E_1$, but instead that simplified function times $e^{-x}$.






      share|cite|improve this answer









      $endgroup$


















        0












        $begingroup$

        If there was a known simplification, we would not be using the function $E_1$, but instead that simplified function times $e^{-x}$.






        share|cite|improve this answer









        $endgroup$
















          0












          0








          0





          $begingroup$

          If there was a known simplification, we would not be using the function $E_1$, but instead that simplified function times $e^{-x}$.






          share|cite|improve this answer









          $endgroup$



          If there was a known simplification, we would not be using the function $E_1$, but instead that simplified function times $e^{-x}$.







          share|cite|improve this answer












          share|cite|improve this answer



          share|cite|improve this answer










          answered Dec 7 '18 at 16:14









          Yves DaoustYves Daoust

          124k671222




          124k671222























              -1












              $begingroup$

              The only operation in the way to simplify this is integrating by parts the $E_1$:



              $$E_1left( xright) = int_x^infty dleft(e^{-t}lntright) + int_x^infty e^{-t}lnt dt = -e^{-x}lnx + int_x^infty e^{-t}lnt dt$$



              Then:



              $$e^xE_1left( x right) = -lnx + e^xint_x^infty e^{-t}lnt dt$$






              share|cite|improve this answer











              $endgroup$


















                -1












                $begingroup$

                The only operation in the way to simplify this is integrating by parts the $E_1$:



                $$E_1left( xright) = int_x^infty dleft(e^{-t}lntright) + int_x^infty e^{-t}lnt dt = -e^{-x}lnx + int_x^infty e^{-t}lnt dt$$



                Then:



                $$e^xE_1left( x right) = -lnx + e^xint_x^infty e^{-t}lnt dt$$






                share|cite|improve this answer











                $endgroup$
















                  -1












                  -1








                  -1





                  $begingroup$

                  The only operation in the way to simplify this is integrating by parts the $E_1$:



                  $$E_1left( xright) = int_x^infty dleft(e^{-t}lntright) + int_x^infty e^{-t}lnt dt = -e^{-x}lnx + int_x^infty e^{-t}lnt dt$$



                  Then:



                  $$e^xE_1left( x right) = -lnx + e^xint_x^infty e^{-t}lnt dt$$






                  share|cite|improve this answer











                  $endgroup$



                  The only operation in the way to simplify this is integrating by parts the $E_1$:



                  $$E_1left( xright) = int_x^infty dleft(e^{-t}lntright) + int_x^infty e^{-t}lnt dt = -e^{-x}lnx + int_x^infty e^{-t}lnt dt$$



                  Then:



                  $$e^xE_1left( x right) = -lnx + e^xint_x^infty e^{-t}lnt dt$$







                  share|cite|improve this answer














                  share|cite|improve this answer



                  share|cite|improve this answer








                  edited Dec 7 '18 at 16:10

























                  answered Dec 5 '18 at 15:06









                  Enrique RenéEnrique René

                  286




                  286






























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