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From a physics problem I'm interested by a closed form of this integral : $$\int_{-\infty}^{+\infty} \frac{1}{\sqrt{P(x)}}e^{-ax^2 - bx - c} dx$$

where $P(x) = \lambda_6 x^6 + ... + \lambda_0$

I just know that $\int_{-\infty}^{+\infty} e^{-ax^2 - bx} dx = \sqrt{\frac{\pi}{a}}e^{\frac{b^2}{4a}}$ but here I have no ideas. Thanks for any suggestions.

Edit : more details for $P(x)$ : $P(x) = \Big(1+(\frac{x+c_1}{c_2})^2\Big)\Big(1+(\frac{x-c_1}{c_2})^2\Big)\Big(\lambda_1[1+(\frac{x-c_3}{c_4})^2] + \lambda_2[1+(\frac{x+c_3}{c_4})^2]\Big)$ with $c_1,c_2,c_3,c_4,\lambda_1, \lambda_2 \in \mathbb R$

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    $\begingroup$ don't you know anything more about the P(x), e.g, roots, ... $\endgroup$ – Cardinal Jul 21 '15 at 9:18
  • $\begingroup$ @Cardinal : I know more about this polynomial but it's a very ugly expression. I will edit my question and put the total expression. $\endgroup$ – user4422 Jul 21 '15 at 9:32
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    $\begingroup$ $P$ is of order 6 ... $\endgroup$ – Math-fun Jul 21 '15 at 9:45
  • $\begingroup$ Stupid mistake, sorry ! First it was a 8 degree polynomial but I had some simplifications. Do you think there is a solution ? $\endgroup$ – user4422 Jul 21 '15 at 9:47
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    $\begingroup$ I set $c_1=c3=0$ and $c_2=c_4=1$ and $a=1,b=c=0$ then you could write your integral as $$\sqrt{\frac{1}{2}}\int_0^{\infty}e^{-x}x^{-1/2}(1+x)^{-3/2}dx$$ I am not sure if this has a closed form solution .. $\endgroup$ – Math-fun Jul 21 '15 at 9:53

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