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The integral is: $$ I =\int_{-\infty}^{+\infty}e^{-\alpha x^{2}}ch(\beta x)dx $$

My investigation gives me: $$ ch(\beta x)=\frac{e^{\beta x}+e^{-\beta x}}{2}$$ $$I= \int_{-\infty}^{+\infty}e^{-\alpha x^{2}}(\frac{e^{\beta x}+e^{-\beta x}}{2})dx = \frac{1}{2}(\int_{-\infty}^{+\infty}e^{-\alpha x^{2}}e^{\beta x}dx + \int_{-\infty}^{+\infty}e^{-\alpha x^{2}}e^{-\beta x}dx)$$

$$\int_{-\infty}^{+\infty}e^{-\alpha x^{2}}e^{\beta x}dx = \int_{-\infty}^{+\infty}e^{-\alpha x^{2}+\beta x}dx$$ I think that the Gaussian Integral is the key. But I can't find the right substitution for $ -\alpha x^{2}+\beta x $ and $-\alpha x^{2}-\beta x$. Any help would be much appreciated!

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    Compete the square. – 2017-02-16

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Hint:

$$-\alpha x^2+\beta x=-\alpha\left(x-\frac\beta{2\alpha}\right)^2+\frac{\beta^2}{4\alpha}$$

Thus, with $u=x-\frac\beta{2\alpha}$, we get

$$\int_{-\infty}^{+\infty}e^{-\alpha x^2+\beta x}\ dx=e^{\frac{\beta^2}{4\alpha}}\int_{-\infty}^{+\infty}e^{-\alpha u^2}\ du$$

For the second integral,

$$-\alpha x^2-\beta x=-\alpha\left(x+\frac\beta{2\alpha}\right)^2+\frac{\beta^2}{4\alpha}$$