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Question-130876




Question Number 130876 by 0731619177 last updated on 30/Jan/21
Answered by Dwaipayan Shikari last updated on 30/Jan/21
∫_(−∞) ^∞ re^(−a^2 (r−(6/a^2 ))^2 +((36)/a^2 )) dr  =e^((36)/a^2 ) ∫_(−∞) ^∞ (z+(6/a^2 ))e^(−a^2 z) dz  =(1/a)e^((36)/a^2 ) +(6/a^2 )e^((36)/a^2 ) (√π)
$$\int_{−\infty} ^{\infty} {re}^{−{a}^{\mathrm{2}} \left({r}−\frac{\mathrm{6}}{{a}^{\mathrm{2}} }\right)^{\mathrm{2}} +\frac{\mathrm{36}}{{a}^{\mathrm{2}} }} {dr} \\ $$$$={e}^{\frac{\mathrm{36}}{{a}^{\mathrm{2}} }} \int_{−\infty} ^{\infty} \left({z}+\frac{\mathrm{6}}{{a}^{\mathrm{2}} }\right){e}^{−{a}^{\mathrm{2}} {z}} {dz} \\ $$$$=\frac{\mathrm{1}}{{a}}{e}^{\frac{\mathrm{36}}{{a}^{\mathrm{2}} }} +\frac{\mathrm{6}}{{a}^{\mathrm{2}} }{e}^{\frac{\mathrm{36}}{{a}^{\mathrm{2}} }} \sqrt{\pi} \\ $$

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