Integrals with the special functions $Ci(x)$ and $erf(x)$












2












$begingroup$


I'm looking for the solutions of the following two integrals:



$$I_1=intlimits_0^infty dx, e^{-x^2}text{Ci}(ax)$$
and
$$I_2=intlimits_0^infty dx, e^{-ax}text{erf}(x)$$
with
$$text{Ci}(x)=intlimits_infty^xfrac{cos(t)}{t}dt$$
and
$$text{erf}(x)=frac{2}{sqrt{pi}}intlimits_0^xe^{-t^2}dt$$



Now I'm not 100% sure what is meant by "the solution of the integrals" since these will probably be not-evaluable. But I'm guessing that the question is to reduce the expression to one single special function in stead of the integral of a special function with an elementary function.



Mathematica yields me the answers:
$$I_1=-frac{sqrt{pi}}{4}Gammaleft(0,frac{a^2}{4}right)$$
and
$$I_2=expleft(frac{a^2}{4}right)frac{1-text{erf}(a/2)}{a}$$



A good first step for evaluating these integrals $I_1$ and $I_2$ seemed to fill in the integral representations of these special functions and try to switch te integrals over $x$ and $t$. However this has not (yet) been a success. I also tried to find a differential equation for these integrals, but also this was not so easy to do. Are there any tips/tricks to evaluate these integrals ?






calculus real-analysis integration definite-integrals special-functions







share|cite|improve this question











$endgroup$

















    2












    $begingroup$


    I'm looking for the solutions of the following two integrals:



    $$I_1=intlimits_0^infty dx, e^{-x^2}text{Ci}(ax)$$
    and
    $$I_2=intlimits_0^infty dx, e^{-ax}text{erf}(x)$$
    with
    $$text{Ci}(x)=intlimits_infty^xfrac{cos(t)}{t}dt$$
    and
    $$text{erf}(x)=frac{2}{sqrt{pi}}intlimits_0^xe^{-t^2}dt$$



    Now I'm not 100% sure what is meant by "the solution of the integrals" since these will probably be not-evaluable. But I'm guessing that the question is to reduce the expression to one single special function in stead of the integral of a special function with an elementary function.



    Mathematica yields me the answers:
    $$I_1=-frac{sqrt{pi}}{4}Gammaleft(0,frac{a^2}{4}right)$$
    and
    $$I_2=expleft(frac{a^2}{4}right)frac{1-text{erf}(a/2)}{a}$$



    A good first step for evaluating these integrals $I_1$ and $I_2$ seemed to fill in the integral representations of these special functions and try to switch te integrals over $x$ and $t$. However this has not (yet) been a success. I also tried to find a differential equation for these integrals, but also this was not so easy to do. Are there any tips/tricks to evaluate these integrals ?






    calculus real-analysis integration definite-integrals special-functions







    share|cite|improve this question











    $endgroup$















      2












      2








      2


      1



      $begingroup$


      I'm looking for the solutions of the following two integrals:



      $$I_1=intlimits_0^infty dx, e^{-x^2}text{Ci}(ax)$$
      and
      $$I_2=intlimits_0^infty dx, e^{-ax}text{erf}(x)$$
      with
      $$text{Ci}(x)=intlimits_infty^xfrac{cos(t)}{t}dt$$
      and
      $$text{erf}(x)=frac{2}{sqrt{pi}}intlimits_0^xe^{-t^2}dt$$



      Now I'm not 100% sure what is meant by "the solution of the integrals" since these will probably be not-evaluable. But I'm guessing that the question is to reduce the expression to one single special function in stead of the integral of a special function with an elementary function.



      Mathematica yields me the answers:
      $$I_1=-frac{sqrt{pi}}{4}Gammaleft(0,frac{a^2}{4}right)$$
      and
      $$I_2=expleft(frac{a^2}{4}right)frac{1-text{erf}(a/2)}{a}$$



      A good first step for evaluating these integrals $I_1$ and $I_2$ seemed to fill in the integral representations of these special functions and try to switch te integrals over $x$ and $t$. However this has not (yet) been a success. I also tried to find a differential equation for these integrals, but also this was not so easy to do. Are there any tips/tricks to evaluate these integrals ?






      calculus real-analysis integration definite-integrals special-functions







      share|cite|improve this question











      $endgroup$




      I'm looking for the solutions of the following two integrals:



      $$I_1=intlimits_0^infty dx, e^{-x^2}text{Ci}(ax)$$
      and
      $$I_2=intlimits_0^infty dx, e^{-ax}text{erf}(x)$$
      with
      $$text{Ci}(x)=intlimits_infty^xfrac{cos(t)}{t}dt$$
      and
      $$text{erf}(x)=frac{2}{sqrt{pi}}intlimits_0^xe^{-t^2}dt$$



      Now I'm not 100% sure what is meant by "the solution of the integrals" since these will probably be not-evaluable. But I'm guessing that the question is to reduce the expression to one single special function in stead of the integral of a special function with an elementary function.



      Mathematica yields me the answers:
      $$I_1=-frac{sqrt{pi}}{4}Gammaleft(0,frac{a^2}{4}right)$$
      and
      $$I_2=expleft(frac{a^2}{4}right)frac{1-text{erf}(a/2)}{a}$$



      A good first step for evaluating these integrals $I_1$ and $I_2$ seemed to fill in the integral representations of these special functions and try to switch te integrals over $x$ and $t$. However this has not (yet) been a success. I also tried to find a differential equation for these integrals, but also this was not so easy to do. Are there any tips/tricks to evaluate these integrals ?






      calculus real-analysis integration definite-integrals special-functions




      calculus real-analysis integration definite-integrals special-functions






      share|cite|improve this question















      share|cite|improve this question













      share|cite|improve this question




      share|cite|improve this question








      edited Jan 15 at 2:30









      Henry Lee

      2,181319




      2,181319










      asked Mar 23 '15 at 12:20









      NickNick

      499518




      499518






















          2 Answers
          2






          active

          oldest

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          3












          $begingroup$

          Hint: Differentiate $I_1(a)$ with regard to a. Similarly, define $~I_2(b)~=~displaystyleint_0^infty e^{-ax}~text{erf}(bx)~dx,~$ and then differentiate it with regard to b.






          share|cite|improve this answer









          $endgroup$













          • $begingroup$
            Nice, adding a factor in order to eliminate one of the integrations !
            $endgroup$
            – Nick
            Mar 23 '15 at 14:51



















          3












          $begingroup$

          For $I_2$, please note that this integral is just the Laplace transform of $text{Erf}(x)$ w.r.t. to the variable $a$



          Furthermore the Laplace transform of an integral $mathcal{L}(int f(x) dx)$ is just $mathcal{L}(f(x))/a$



          Using this we find that



          $$
          I_2=frac{2}{asqrt{pi}}int_0^{infty}e^{- a x}e^{-x^2}dx
          $$



          Completing the square in the exponent and changing variables $y=frac{a}{2}+x$



          This turns into



          $$
          I_2=2frac{ e^{frac{a^2}{4}}}{asqrt{pi}}int_{frac{a}{2}}^{infty}e^{-y^2}dy
          $$



          Now using straightforwardly the definition of the Errorfunction we obtain



          $$
          I_2=frac{e^{frac{a^2}{4}}}{a}left(1-text{Erf}(a/2)right)
          $$



          As promised!



          I will have a look on $I_1$ later






          share|cite|improve this answer









          $endgroup$













          • $begingroup$
            thanks a lot, that's already a nice first step. I somehow have the feeling that both integrals $I_1$ and $I_2$ should be analogous (due to a slight analogy in the answer). So I didn't consider Laplace because it does not appear in the first integral.
            $endgroup$
            – Nick
            Mar 23 '15 at 13:27











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          2 Answers
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          2 Answers
          2






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          active

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          3












          $begingroup$

          Hint: Differentiate $I_1(a)$ with regard to a. Similarly, define $~I_2(b)~=~displaystyleint_0^infty e^{-ax}~text{erf}(bx)~dx,~$ and then differentiate it with regard to b.






          share|cite|improve this answer









          $endgroup$













          • $begingroup$
            Nice, adding a factor in order to eliminate one of the integrations !
            $endgroup$
            – Nick
            Mar 23 '15 at 14:51
















          3












          $begingroup$

          Hint: Differentiate $I_1(a)$ with regard to a. Similarly, define $~I_2(b)~=~displaystyleint_0^infty e^{-ax}~text{erf}(bx)~dx,~$ and then differentiate it with regard to b.






          share|cite|improve this answer









          $endgroup$













          • $begingroup$
            Nice, adding a factor in order to eliminate one of the integrations !
            $endgroup$
            – Nick
            Mar 23 '15 at 14:51














          3












          3








          3





          $begingroup$

          Hint: Differentiate $I_1(a)$ with regard to a. Similarly, define $~I_2(b)~=~displaystyleint_0^infty e^{-ax}~text{erf}(bx)~dx,~$ and then differentiate it with regard to b.






          share|cite|improve this answer









          $endgroup$



          Hint: Differentiate $I_1(a)$ with regard to a. Similarly, define $~I_2(b)~=~displaystyleint_0^infty e^{-ax}~text{erf}(bx)~dx,~$ and then differentiate it with regard to b.







          share|cite|improve this answer












          share|cite|improve this answer



          share|cite|improve this answer










          answered Mar 23 '15 at 13:27









          LucianLucian

          41.5k159131




          41.5k159131












          • $begingroup$
            Nice, adding a factor in order to eliminate one of the integrations !
            $endgroup$
            – Nick
            Mar 23 '15 at 14:51


















          • $begingroup$
            Nice, adding a factor in order to eliminate one of the integrations !
            $endgroup$
            – Nick
            Mar 23 '15 at 14:51
















          $begingroup$
          Nice, adding a factor in order to eliminate one of the integrations !
          $endgroup$
          – Nick
          Mar 23 '15 at 14:51




          $begingroup$
          Nice, adding a factor in order to eliminate one of the integrations !
          $endgroup$
          – Nick
          Mar 23 '15 at 14:51











          3












          $begingroup$

          For $I_2$, please note that this integral is just the Laplace transform of $text{Erf}(x)$ w.r.t. to the variable $a$



          Furthermore the Laplace transform of an integral $mathcal{L}(int f(x) dx)$ is just $mathcal{L}(f(x))/a$



          Using this we find that



          $$
          I_2=frac{2}{asqrt{pi}}int_0^{infty}e^{- a x}e^{-x^2}dx
          $$



          Completing the square in the exponent and changing variables $y=frac{a}{2}+x$



          This turns into



          $$
          I_2=2frac{ e^{frac{a^2}{4}}}{asqrt{pi}}int_{frac{a}{2}}^{infty}e^{-y^2}dy
          $$



          Now using straightforwardly the definition of the Errorfunction we obtain



          $$
          I_2=frac{e^{frac{a^2}{4}}}{a}left(1-text{Erf}(a/2)right)
          $$



          As promised!



          I will have a look on $I_1$ later






          share|cite|improve this answer









          $endgroup$













          • $begingroup$
            thanks a lot, that's already a nice first step. I somehow have the feeling that both integrals $I_1$ and $I_2$ should be analogous (due to a slight analogy in the answer). So I didn't consider Laplace because it does not appear in the first integral.
            $endgroup$
            – Nick
            Mar 23 '15 at 13:27
















          3












          $begingroup$

          For $I_2$, please note that this integral is just the Laplace transform of $text{Erf}(x)$ w.r.t. to the variable $a$



          Furthermore the Laplace transform of an integral $mathcal{L}(int f(x) dx)$ is just $mathcal{L}(f(x))/a$



          Using this we find that



          $$
          I_2=frac{2}{asqrt{pi}}int_0^{infty}e^{- a x}e^{-x^2}dx
          $$



          Completing the square in the exponent and changing variables $y=frac{a}{2}+x$



          This turns into



          $$
          I_2=2frac{ e^{frac{a^2}{4}}}{asqrt{pi}}int_{frac{a}{2}}^{infty}e^{-y^2}dy
          $$



          Now using straightforwardly the definition of the Errorfunction we obtain



          $$
          I_2=frac{e^{frac{a^2}{4}}}{a}left(1-text{Erf}(a/2)right)
          $$



          As promised!



          I will have a look on $I_1$ later






          share|cite|improve this answer









          $endgroup$













          • $begingroup$
            thanks a lot, that's already a nice first step. I somehow have the feeling that both integrals $I_1$ and $I_2$ should be analogous (due to a slight analogy in the answer). So I didn't consider Laplace because it does not appear in the first integral.
            $endgroup$
            – Nick
            Mar 23 '15 at 13:27














          3












          3








          3





          $begingroup$

          For $I_2$, please note that this integral is just the Laplace transform of $text{Erf}(x)$ w.r.t. to the variable $a$



          Furthermore the Laplace transform of an integral $mathcal{L}(int f(x) dx)$ is just $mathcal{L}(f(x))/a$



          Using this we find that



          $$
          I_2=frac{2}{asqrt{pi}}int_0^{infty}e^{- a x}e^{-x^2}dx
          $$



          Completing the square in the exponent and changing variables $y=frac{a}{2}+x$



          This turns into



          $$
          I_2=2frac{ e^{frac{a^2}{4}}}{asqrt{pi}}int_{frac{a}{2}}^{infty}e^{-y^2}dy
          $$



          Now using straightforwardly the definition of the Errorfunction we obtain



          $$
          I_2=frac{e^{frac{a^2}{4}}}{a}left(1-text{Erf}(a/2)right)
          $$



          As promised!



          I will have a look on $I_1$ later






          share|cite|improve this answer









          $endgroup$



          For $I_2$, please note that this integral is just the Laplace transform of $text{Erf}(x)$ w.r.t. to the variable $a$



          Furthermore the Laplace transform of an integral $mathcal{L}(int f(x) dx)$ is just $mathcal{L}(f(x))/a$



          Using this we find that



          $$
          I_2=frac{2}{asqrt{pi}}int_0^{infty}e^{- a x}e^{-x^2}dx
          $$



          Completing the square in the exponent and changing variables $y=frac{a}{2}+x$



          This turns into



          $$
          I_2=2frac{ e^{frac{a^2}{4}}}{asqrt{pi}}int_{frac{a}{2}}^{infty}e^{-y^2}dy
          $$



          Now using straightforwardly the definition of the Errorfunction we obtain



          $$
          I_2=frac{e^{frac{a^2}{4}}}{a}left(1-text{Erf}(a/2)right)
          $$



          As promised!



          I will have a look on $I_1$ later







          share|cite|improve this answer












          share|cite|improve this answer



          share|cite|improve this answer










          answered Mar 23 '15 at 13:24









          tiredtired

          10.6k12044




          10.6k12044












          • $begingroup$
            thanks a lot, that's already a nice first step. I somehow have the feeling that both integrals $I_1$ and $I_2$ should be analogous (due to a slight analogy in the answer). So I didn't consider Laplace because it does not appear in the first integral.
            $endgroup$
            – Nick
            Mar 23 '15 at 13:27


















          • $begingroup$
            thanks a lot, that's already a nice first step. I somehow have the feeling that both integrals $I_1$ and $I_2$ should be analogous (due to a slight analogy in the answer). So I didn't consider Laplace because it does not appear in the first integral.
            $endgroup$
            – Nick
            Mar 23 '15 at 13:27
















          $begingroup$
          thanks a lot, that's already a nice first step. I somehow have the feeling that both integrals $I_1$ and $I_2$ should be analogous (due to a slight analogy in the answer). So I didn't consider Laplace because it does not appear in the first integral.
          $endgroup$
          – Nick
          Mar 23 '15 at 13:27




          $begingroup$
          thanks a lot, that's already a nice first step. I somehow have the feeling that both integrals $I_1$ and $I_2$ should be analogous (due to a slight analogy in the answer). So I didn't consider Laplace because it does not appear in the first integral.
          $endgroup$
          – Nick
          Mar 23 '15 at 13:27


















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