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

How to find integral:

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

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.
@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