Computation of a double exponential integral












1












$begingroup$


I want to understand the behavior of this integral
$$ int_0^x e^{-frac{c}{t^2}} frac{1}{t^5} e^{-frac{c_2}{(x-t)^2}} frac{1}{(x-t)^5} dt. $$
The ideal answer would be a way to explicitly compute this, but I have no clue. This integral is clearly definite, but my problem is to understand quantitatively its behavior for small and large values of $c$ and $c_2$. Any idea on how to proceed?






analysis definite-integrals







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  • $begingroup$
    At first i would simplify the integrand!
    $endgroup$
    – Dr. Sonnhard Graubner
    Jan 19 at 12:32










  • $begingroup$
    What would be the best way to do so? The two obvious changes of variable ($t to frac{1}{t}$ or $t to frac{1}{x-t}$) seem to lead to something more complicated.
    $endgroup$
    – Gâteau-Gallois
    Jan 19 at 12:44












  • $begingroup$
    Why did you delete the second part of the question ?
    $endgroup$
    – Claude Leibovici
    Jan 19 at 13:14










  • $begingroup$
    I think the integral I had was not convergent, near 0 it is homogeneous to $1/t^k$ for $k geq 2$... I need to check again the entire series computation
    $endgroup$
    – Gâteau-Gallois
    Jan 19 at 13:38
















1












$begingroup$


I want to understand the behavior of this integral
$$ int_0^x e^{-frac{c}{t^2}} frac{1}{t^5} e^{-frac{c_2}{(x-t)^2}} frac{1}{(x-t)^5} dt. $$
The ideal answer would be a way to explicitly compute this, but I have no clue. This integral is clearly definite, but my problem is to understand quantitatively its behavior for small and large values of $c$ and $c_2$. Any idea on how to proceed?






analysis definite-integrals







share|cite|improve this question











$endgroup$












  • $begingroup$
    At first i would simplify the integrand!
    $endgroup$
    – Dr. Sonnhard Graubner
    Jan 19 at 12:32










  • $begingroup$
    What would be the best way to do so? The two obvious changes of variable ($t to frac{1}{t}$ or $t to frac{1}{x-t}$) seem to lead to something more complicated.
    $endgroup$
    – Gâteau-Gallois
    Jan 19 at 12:44












  • $begingroup$
    Why did you delete the second part of the question ?
    $endgroup$
    – Claude Leibovici
    Jan 19 at 13:14










  • $begingroup$
    I think the integral I had was not convergent, near 0 it is homogeneous to $1/t^k$ for $k geq 2$... I need to check again the entire series computation
    $endgroup$
    – Gâteau-Gallois
    Jan 19 at 13:38














1












1








1





$begingroup$


I want to understand the behavior of this integral
$$ int_0^x e^{-frac{c}{t^2}} frac{1}{t^5} e^{-frac{c_2}{(x-t)^2}} frac{1}{(x-t)^5} dt. $$
The ideal answer would be a way to explicitly compute this, but I have no clue. This integral is clearly definite, but my problem is to understand quantitatively its behavior for small and large values of $c$ and $c_2$. Any idea on how to proceed?






analysis definite-integrals







share|cite|improve this question











$endgroup$




I want to understand the behavior of this integral
$$ int_0^x e^{-frac{c}{t^2}} frac{1}{t^5} e^{-frac{c_2}{(x-t)^2}} frac{1}{(x-t)^5} dt. $$
The ideal answer would be a way to explicitly compute this, but I have no clue. This integral is clearly definite, but my problem is to understand quantitatively its behavior for small and large values of $c$ and $c_2$. Any idea on how to proceed?






analysis definite-integrals




analysis definite-integrals






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share|cite|improve this question













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edited Jan 19 at 13:00







Gâteau-Gallois

















asked Jan 19 at 12:30









Gâteau-GalloisGâteau-Gallois

362113




362113












  • $begingroup$
    At first i would simplify the integrand!
    $endgroup$
    – Dr. Sonnhard Graubner
    Jan 19 at 12:32










  • $begingroup$
    What would be the best way to do so? The two obvious changes of variable ($t to frac{1}{t}$ or $t to frac{1}{x-t}$) seem to lead to something more complicated.
    $endgroup$
    – Gâteau-Gallois
    Jan 19 at 12:44












  • $begingroup$
    Why did you delete the second part of the question ?
    $endgroup$
    – Claude Leibovici
    Jan 19 at 13:14










  • $begingroup$
    I think the integral I had was not convergent, near 0 it is homogeneous to $1/t^k$ for $k geq 2$... I need to check again the entire series computation
    $endgroup$
    – Gâteau-Gallois
    Jan 19 at 13:38


















  • $begingroup$
    At first i would simplify the integrand!
    $endgroup$
    – Dr. Sonnhard Graubner
    Jan 19 at 12:32










  • $begingroup$
    What would be the best way to do so? The two obvious changes of variable ($t to frac{1}{t}$ or $t to frac{1}{x-t}$) seem to lead to something more complicated.
    $endgroup$
    – Gâteau-Gallois
    Jan 19 at 12:44












  • $begingroup$
    Why did you delete the second part of the question ?
    $endgroup$
    – Claude Leibovici
    Jan 19 at 13:14










  • $begingroup$
    I think the integral I had was not convergent, near 0 it is homogeneous to $1/t^k$ for $k geq 2$... I need to check again the entire series computation
    $endgroup$
    – Gâteau-Gallois
    Jan 19 at 13:38
















$begingroup$
At first i would simplify the integrand!
$endgroup$
– Dr. Sonnhard Graubner
Jan 19 at 12:32




$begingroup$
At first i would simplify the integrand!
$endgroup$
– Dr. Sonnhard Graubner
Jan 19 at 12:32












$begingroup$
What would be the best way to do so? The two obvious changes of variable ($t to frac{1}{t}$ or $t to frac{1}{x-t}$) seem to lead to something more complicated.
$endgroup$
– Gâteau-Gallois
Jan 19 at 12:44






$begingroup$
What would be the best way to do so? The two obvious changes of variable ($t to frac{1}{t}$ or $t to frac{1}{x-t}$) seem to lead to something more complicated.
$endgroup$
– Gâteau-Gallois
Jan 19 at 12:44














$begingroup$
Why did you delete the second part of the question ?
$endgroup$
– Claude Leibovici
Jan 19 at 13:14




$begingroup$
Why did you delete the second part of the question ?
$endgroup$
– Claude Leibovici
Jan 19 at 13:14












$begingroup$
I think the integral I had was not convergent, near 0 it is homogeneous to $1/t^k$ for $k geq 2$... I need to check again the entire series computation
$endgroup$
– Gâteau-Gallois
Jan 19 at 13:38




$begingroup$
I think the integral I had was not convergent, near 0 it is homogeneous to $1/t^k$ for $k geq 2$... I need to check again the entire series computation
$endgroup$
– Gâteau-Gallois
Jan 19 at 13:38










2 Answers
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$begingroup$

Hint



Assuming $t<x$, $x>0$,$$intfrac{1}{t^k} frac{1}{(x-t)^l},dt=x^{-k-l+1} B_{frac{t}{x}}(1-k,1-l)$$






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    0












    $begingroup$

    I mean this here $$e^{-frac{c}{t^2}-frac{c_2}{(x-t)^2}}cdot frac{1}{(t(x-t))^5}$$






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      2 Answers
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      2 Answers
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      2












      $begingroup$

      Hint



      Assuming $t<x$, $x>0$,$$intfrac{1}{t^k} frac{1}{(x-t)^l},dt=x^{-k-l+1} B_{frac{t}{x}}(1-k,1-l)$$






      share|cite|improve this answer









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        2












        $begingroup$

        Hint



        Assuming $t<x$, $x>0$,$$intfrac{1}{t^k} frac{1}{(x-t)^l},dt=x^{-k-l+1} B_{frac{t}{x}}(1-k,1-l)$$






        share|cite|improve this answer









        $endgroup$
















          2












          2








          2





          $begingroup$

          Hint



          Assuming $t<x$, $x>0$,$$intfrac{1}{t^k} frac{1}{(x-t)^l},dt=x^{-k-l+1} B_{frac{t}{x}}(1-k,1-l)$$






          share|cite|improve this answer









          $endgroup$



          Hint



          Assuming $t<x$, $x>0$,$$intfrac{1}{t^k} frac{1}{(x-t)^l},dt=x^{-k-l+1} B_{frac{t}{x}}(1-k,1-l)$$







          share|cite|improve this answer












          share|cite|improve this answer



          share|cite|improve this answer










          answered Jan 19 at 13:09









          Claude LeiboviciClaude Leibovici

          123k1157134




          123k1157134























              0












              $begingroup$

              I mean this here $$e^{-frac{c}{t^2}-frac{c_2}{(x-t)^2}}cdot frac{1}{(t(x-t))^5}$$






              share|cite|improve this answer









              $endgroup$


















                0












                $begingroup$

                I mean this here $$e^{-frac{c}{t^2}-frac{c_2}{(x-t)^2}}cdot frac{1}{(t(x-t))^5}$$






                share|cite|improve this answer









                $endgroup$
















                  0












                  0








                  0





                  $begingroup$

                  I mean this here $$e^{-frac{c}{t^2}-frac{c_2}{(x-t)^2}}cdot frac{1}{(t(x-t))^5}$$






                  share|cite|improve this answer









                  $endgroup$



                  I mean this here $$e^{-frac{c}{t^2}-frac{c_2}{(x-t)^2}}cdot frac{1}{(t(x-t))^5}$$







                  share|cite|improve this answer












                  share|cite|improve this answer



                  share|cite|improve this answer










                  answered Jan 19 at 12:51









                  Dr. Sonnhard GraubnerDr. Sonnhard Graubner

                  76.7k42866




                  76.7k42866






























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