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Question Number 142789 by daniel1301 last updated on 05/Jun/21

$$\int e^{\frac{x^2}{2}}dx=??$$

Answered by MJS_new last updated on 05/Jun/21

$$\int {\mathrm{e}}^{\frac{x^2}{2}}dx=$$$$[t=\frac{\sqrt{2}x}{2}\to dx=\sqrt{2}dt]$$$$=\sqrt{2}\int {\mathrm{e}}^{t^2}dt=\sqrt{2}\times \frac{\sqrt{\pi }}{2}\int {\mathrm{e}}^{t^2}\times \frac{2}{\sqrt{\pi }}dt=\frac{\sqrt{2\pi }}{2}\int \frac{{2\mathrm{e}}^{t^2}}{\sqrt{\pi }}dt=$$$$=\frac{\sqrt{2\pi }}{2}\mathrm{erfi}t=$$$$=\frac{\sqrt{2\pi }}{2}\mathrm{erfi}\frac{\sqrt{2}x}{2}+C(!!)$$

Commented by daniel1301 last updated on 05/Jun/21

$$beautiful$$

Answered by Dwaipayan Shikari last updated on 05/Jun/21

$$x=\sqrt{2}t$$$$\sqrt{2}\int e^{t^2}dt$$$$=\sqrt{2}({te}^{t^2}-2\int t^2{e}^{t^2}dt)$$$$=\sqrt{2}({te}^{t^2}-\frac{2}{3}{e}^{t^2}{t}^3+\frac{2^2}{3}\int t^4{e}^{t^2}dt)$$$$=\sqrt{2}({te}^{t^2}-\frac{2}{3}{e}^{t^2}{t}^3+\frac{2^2}{3.5}{e}^{t^2}{t}^5+\frac{2^3}{3.5.7}{e}^{t^2}{t}^7+...)+C$$$$=\sqrt{2}{e}^{t^2}(t-\frac{2}{3}{t}^3+\frac{2^2}{3.5}{t}^5-\frac{2^3}{3.5.7}{t}^7+...)+C$$$$=e^{x^2/2}(x-\frac{1}{3}{x}^3+\frac{1}{3.5}{x}^5-\frac{1}{3.5.7}{x}^7+...)+C$$$${\int }_0^1{e}^{x^2/2}dx=\sqrt{e}(1-\frac{1}{1.3}+\frac{1}{1.3.5}-\frac{1}{1.3.5.7}+..)$$

Commented by daniel1301 last updated on 05/Jun/21

$$thanks$$

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