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Question Number 203301 by MathematicalUser2357 last updated on 15/Jan/24

$$\mathrm{is}$$$$\int \sqrt{x}{e}^{-x}dx$$$$=\underset{n=1}{\sum ^{\mathrm{\infty }}}\left(\left(\underset{k=1}{\prod ^n}\frac{2}{2k+1}\right)x^n\sqrt{x}{e}^{-x}\right)+C?$$

Answered by witcher3 last updated on 15/Jan/24

$$\int {\mathrm{x}}^{\mathrm{a}}{\mathrm{e}}^{-\mathrm{x}}\mathrm{dx}=\frac{{\mathrm{x}}^{\mathrm{n}+1}}{\mathrm{n}+1}{\mathrm{e}}^{-\mathrm{x}}+\frac{1}{\mathrm{n}+1}\int {\mathrm{x}}^{\mathrm{n}+1}{\mathrm{e}}^{-\mathrm{x}}$$$${\mathrm{I}}_{\mathrm{n}}=\frac{{\mathrm{x}}^{\mathrm{n}+1}{\mathrm{e}}^{-\mathrm{x}}}{\mathrm{n}+1}+\frac{1}{\mathrm{n}+1}{\mathrm{I}}_{\mathrm{n}+1}$$$${\mathrm{I}}_{\mathrm{n}}=\frac{1}{\mathrm{n}+1}({\mathrm{x}}^{\mathrm{n}+1}{\mathrm{e}}^{-\mathrm{x}}+\frac{1}{\mathrm{n}+2}({\mathrm{x}}^{\mathrm{n}+2}{\mathrm{e}}^{-\mathrm{x}}+\frac{1}{\mathrm{n}+3}({\mathrm{x}}^{\mathrm{n}+3}{\mathrm{e}}^{-\mathrm{x}}+\frac{1}{\mathrm{n}+4}{\mathrm{I}}_{\mathrm{n}+4}+....)$$$$\sum _{\mathrm{m}⩾1}\left(\underset{\mathrm{k}=1}{\prod ^{\mathrm{m}}}\frac{1}{(\mathrm{n}+\mathrm{k})}\right){\mathrm{x}}^{\mathrm{n}+\mathrm{m}}{\mathrm{e}}^{-\mathrm{x}}+\mathrm{c}$$$$\int \sqrt{\mathrm{x}}{\mathrm{e}}^{-\mathrm{x}}\mathrm{dx};\mathrm{n}=\frac{1}{2}$$$$=\sum _{\mathrm{n}⩾1}\left(\underset{\mathrm{k}=1}{\prod ^{\mathrm{n}}}\frac{1}{(\frac{1}{2}+\mathrm{k})}\right){\mathrm{x}}^{\frac{1}{2}+\mathrm{n}}{\mathrm{e}}^{-\mathrm{x}}=\sum _{\mathrm{n}⩾1}\left(\underset{\mathrm{k}=1}{\prod ^{\mathrm{n}}}\frac{2}{\mathrm{k}+2}\right){\mathrm{x}}^{\mathrm{n}}\sqrt{\mathrm{x}}{\mathrm{e}}^{-\mathrm{x}}+\mathrm{c}$$$$$$

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