Integral $\int_{ \ 0}^{\ L} \exp{(\frac{-(x-x_0)^2}{4n})\sin({m\pi\over L}(x-A)}\ \mathrm dx$

How to find integral:

$$I=\int_{ \ 0}^{\ L} \exp{\left(\frac{-(x-x_0)^2}{4n}\right)\sin\left({m\pi\over L}(x-A)\right)}dx$$

Thanks in advance.

My try: By substitution I get $$z=(x-x_0)^2$$ $$\sqrt z+x_0=x$$ $${1\over2\sqrt z}dz=dx$$ Finally, get this integral $$I=\int_{ \ 0}^{\ L} \exp{\left(\frac{-z}{4n}\right)\sin\left({m\pi\over L}(\sqrt z+x_0-A)\right)}{1\over2\sqrt z}dz$$ I tried to solve this by parts but I lost.

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What have you tried? –  Alizter Nov 19 '13 at 19:30
@Alizter I did this By substitution I get $$z=(x-x_0)^2$$ $$\sqrt z+x_0=x$$ $${1\over2\sqrt z}dz=dx$$ Finally, get this integral $$I=\int_{ \ 0}^{\ L} \exp{\left(\frac{-z}{4n}\right)\sin\left({m\pi\over L}(\sqrt z+x_0-A)\right)}{1\over2\sqrt z}dz$$ I tried to solve this by parts but I lost. –  Malik Nov 20 '13 at 1:17
Mathematica 9 gives a result for this integral. It's quite long and in terms of the imaginary error function. –  Leonida Nov 20 '13 at 2:17
@NijankowskiV. Can you please tell me how to implement that one in Mathematica? I have never used this before. –  Malik Nov 21 '13 at 2:36