TITLE: how to evaluate $ \int_0^\infty e^{- \left ( x - \frac a x \right )^2} dx $
QUESTION [9 upvotes]: How do I evaluate the following definite integral$$ \int_0^\infty e^{- \left ( x - \frac a x \right )^2} dx $$

REPLY [3 votes]: another solution from me :
$$I=\int_{0}^{\infty }e^{-(x-\frac{a}{x})^2}dx\\
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=\frac{1}{2}\int_{0}^{\infty }(1+\frac{a}{x^2}+(1-\frac{a}{x^2})) .e^{-(x-\frac{a}{x})^2}dx\\$$
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$$=\frac{1}{2}\int_{0}^{\infty }e^{-(x-\frac{a}{x})^2}d(x-\frac{a}{x})+\frac{1}{2}\int_{0}^{\infty }e^{-(x-\frac{a}{x})^2}d(x+\frac{a}{x})\\
\\$$
$$=\frac{1}{2}\int_{0}^{\infty }e^{-(x-\frac{a}{x})^2}d(x-\frac{a}{x})+\frac{e^{4a}}{2}\int_{0}^{\infty }e^{-(x+\frac{a}{x})^2}d(x+\frac{a}{x})\\
\\$$
$$=\frac{1}{2}\int_{-\infty }^{\infty }e^{-y^2}dy+\frac{e^{4a}}{2}\int_{\infty }^{\infty }e^{-y^2}dy\\
\\
\\$$
$$=\int_{0}^{\infty }e^{-y^2}dy+0=\sqrt{\pi }\\
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\therefore \ \ \ \ \ I=\int_{0}^{\infty }e^{-(x-\frac{a}{x})^2}dx=\sqrt{\pi }$$