# Calculate $\int_{0}^{+\infty}\frac{e^{-x^2}}{(x+\frac{1}{2})^2}\mathrm{d}x$ through the Guasssian integral

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.

• – J.G.
Mar 25 '20 at 12:57

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 $$I_1$$.

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