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I'm attempting to evaluate the following integral, so far, with little success. Any help would be appreciated:

$$ \ \int_0^{\infty } \exp\left(-g x-\frac{x^2}{2}-\frac{x^2 z}{1-z}\right) x^k \sin(hx) \, dx $$

All paramaters are real.

Mx

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Why did you write it in Mathematca notation rather than mathematics notation? –  GEdgar Jun 12 '12 at 15:04
    
@GEdgar: Probably familiarity. –  Cameron Buie Jun 12 '12 at 15:13
    
I would consider this as a family of integrals parametrised by $k$. Write $I_k$ for the corresponding integral. Integration by parts should give you a relation between different $I_k$ probably recursion-like. Maybe this helps. –  Simon Markett Jun 12 '12 at 15:16
    
One evaluates integrals, and solves equations. –  Joe Jun 12 '12 at 15:48
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There is a $(1-z)^{-1-k}$ term that seems irrelevant. A typo? –  Ragib Zaman Jun 12 '12 at 16:29

1 Answer 1

up vote 3 down vote accepted

Combining summand is the exponents one has that the integral is a Laplace transform $x\to g$ of the function $f(x,a,h)=e^{-a x^2} x^k \sin hx\;$. For $k=0$, $a,g>0\;$ Mma gives $$ L[f]= -\frac{i \sqrt{\pi } e^{\frac{(g-i h)^2}{4 a}} \left(\text{erfc}\left(\frac{g-i h}{2 \sqrt{a}}\right)-e^{\frac{i g h}{a}} \text{erfc}\left(\frac{g+i h}{2 \sqrt{a}}\right)\right)}{4 \sqrt{a}}. $$ Now note that for natural $k$ the result is equal to $(-1)^k\frac{\partial^k}{\partial g^k}L[f]$, so it seems where is no good formula for arbitrary $k$.

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