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Question Number 204707 by Mummyjay last updated on 25/Feb/24
integrate ∫_0 ^∞ (e^(−x^2 ) /(1+e^x ))dx
$$\boldsymbol{{integrate}}\:\int_{\mathrm{0}} ^{\infty} \frac{\boldsymbol{{e}}^{−\boldsymbol{{x}}^{\mathrm{2}} } }{\mathrm{1}+\boldsymbol{{e}}^{\boldsymbol{{x}}} }\boldsymbol{{dx}} \\ $$
Answered by witcher3 last updated on 26/Feb/24
Ω=∫_0 ^∞ (e^(−x^2 −x) /(1+e^(−x) ))dx=Σ_(n≥0) (−1)^n ∫_0 ^∞ e^(−(x^2 +(1+n)x)) dx x=(1+n)y =Σ_(n≥0) (−1)^n (n+1)∫_0 ^∞ e^(−(1+n)^2 ((y+(1/2))^2 −(1/4))) dy;(1+n)(y+(1/2))=z =Σ_(n≥0) (−1)^n e^((((1+n)/2))^2 ) ∫_((n+1)/2) ^∞ e^(−z^2 ) dz (2/( (√π)))∫_0 ^x e^(−t^2 ) dt.erfc(x) =Σ_(n≥0) (−1)^n e^((((1+n)/2))^2 ) .(((√π)/2)−((√π)/2)erfc(((n+1)/2))) =((√π)/2)Σ_(n≥0) (−1)^n e^((((1+n)/2))^2 ) (1−erfc(((n+1)/2)))
$$\Omega=\int_{\mathrm{0}} ^{\infty} \frac{\mathrm{e}^{−\mathrm{x}^{\mathrm{2}} −\mathrm{x}} }{\mathrm{1}+\mathrm{e}^{−\mathrm{x}} }\mathrm{dx}=\underset{\mathrm{n}\geqslant\mathrm{0}} {\sum}\left(−\mathrm{1}\right)^{\mathrm{n}} \int_{\mathrm{0}} ^{\infty} \mathrm{e}^{−\left(\mathrm{x}^{\mathrm{2}} +\left(\mathrm{1}+\mathrm{n}\right)\mathrm{x}\right)} \mathrm{dx} \\ $$$$\mathrm{x}=\left(\mathrm{1}+\mathrm{n}\right)\mathrm{y} \\ $$$$=\underset{\mathrm{n}\geqslant\mathrm{0}} {\sum}\left(−\mathrm{1}\right)^{\mathrm{n}} \left(\mathrm{n}+\mathrm{1}\right)\int_{\mathrm{0}} ^{\infty} \mathrm{e}^{−\left(\mathrm{1}+\mathrm{n}\right)^{\mathrm{2}} \left(\left(\mathrm{y}+\frac{\mathrm{1}}{\mathrm{2}}\right)^{\mathrm{2}} −\frac{\mathrm{1}}{\mathrm{4}}\right)} \mathrm{dy};\left(\mathrm{1}+\mathrm{n}\right)\left(\mathrm{y}+\frac{\mathrm{1}}{\mathrm{2}}\right)=\mathrm{z} \\ $$$$=\underset{\mathrm{n}\geqslant\mathrm{0}} {\sum}\left(−\mathrm{1}\right)^{\mathrm{n}} \mathrm{e}^{\left(\frac{\mathrm{1}+\mathrm{n}}{\mathrm{2}}\right)^{\mathrm{2}} } \int_{\frac{\mathrm{n}+\mathrm{1}}{\mathrm{2}}} ^{\infty} \mathrm{e}^{−\mathrm{z}^{\mathrm{2}} } \mathrm{dz} \\ $$$$\frac{\mathrm{2}}{\:\sqrt{\pi}}\int_{\mathrm{0}} ^{\mathrm{x}} \mathrm{e}^{−\mathrm{t}^{\mathrm{2}} } \mathrm{dt}.\mathrm{erfc}\left(\mathrm{x}\right) \\ $$$$=\underset{\mathrm{n}\geqslant\mathrm{0}} {\sum}\left(−\mathrm{1}\right)^{\mathrm{n}} \mathrm{e}^{\left(\frac{\mathrm{1}+\mathrm{n}}{\mathrm{2}}\right)^{\mathrm{2}} } .\left(\frac{\sqrt{\pi}}{\mathrm{2}}−\frac{\sqrt{\pi}}{\mathrm{2}}\mathrm{erfc}\left(\frac{\mathrm{n}+\mathrm{1}}{\mathrm{2}}\right)\right) \\ $$$$=\frac{\sqrt{\pi}}{\mathrm{2}}\underset{\mathrm{n}\geqslant\mathrm{0}} {\sum}\left(−\mathrm{1}\right)^{\mathrm{n}} \mathrm{e}^{\left(\frac{\mathrm{1}+\mathrm{n}}{\mathrm{2}}\right)^{\mathrm{2}} } \left(\mathrm{1}−\mathrm{erfc}\left(\frac{\mathrm{n}+\mathrm{1}}{\mathrm{2}}\right)\right) \\ $$

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