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The problem is to determine the following integral: $$\int_{0}^{+\infty}\frac{e^{-x^2}}{(x+\frac{1}{2})^2}\mathrm{d}x$$ through the Guassian integral: $$\int_{0}^{+\infty}e^{-x^2}\mathrm{d}x=\frac{\sqrt{\pi}}{2}.$$

All that has occurred to me is integration by parts, which gives $$\int_{0}^{+\infty}e^{-x^2}\mathrm{d}\frac{-1}{x+\frac{1}{2}}=\int_{0}^{+\infty}\frac{e^{-x^2}}{x+\frac{1}{2}}\mathrm{d}x+2-\sqrt{\pi}.$$ However, it still seems far from the result we're looking for, and I get stuck. Please help.

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Consider $I = \int_0^\infty \frac{e^{-x^2}}{(x+\frac{1}{2})^2} dx$ and $I_m = \int_0^\infty x^me^{-x^2}dx$ for non-negative integer m.Using taylor series tell us: $(x+\frac{1}{2})^{-2}=\sum_{m=0}^\infty(-1)^m(m+1)2^{m+2}x^m$ so we have: $$I = \sum_{m=0}^\infty(-1)^m(m+1)2^{m+2}\int_0^\infty x^me^{-x^2}dx$$On the other hand use integration by part for $I_m$ we have: $I_m=\frac{1}{2}(m-1)I_{m-2}$ and $I_1 = \frac{1}{2}$ and $I_0=\frac{\sqrt{\pi}}{2}$.

Now you can write $$I=\sum_{m=0}^∞(−1)^m(m+1)2^{m+2}I_m$$Also $$I_m = (\frac{1}{2})^{[\frac{m}{2}]}I_s\prod_{k=0}^{[\frac{m}{2}]}(m-(2k+1))$$ That s is 1 if m is odd and zero if m is even. It Means that we have written I using $I_0$ and $I_1$.

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    $\begingroup$ Thanks for your answer, but here I wonder whether applying Taylor series here is rigorous. Can you show rigidity? $\endgroup$
    – Juggler
    Mar 25 '20 at 13:19

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