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Question Number 144849 by Dwaipayan Shikari last updated on 29/Jun/21

$${\int }_0^2\frac{1}{e^{{\left\{x\right\}}^2}+1}dx\left\{x\right\}isfractionalpartofx$$

Answered by mathmax by abdo last updated on 29/Jun/21

$$\mathrm{\Phi }={\int }_0^2\frac{\mathrm{dx}}{{\mathrm{e}}^{{\left\{\mathrm{x}\right\}}^2}+1}\mathrm{we}\mathrm{have}\mathrm{x}=\left[\mathrm{x}\right]+\left\{\mathrm{x}\right\}\Rightarrow \left\{\mathrm{x}\right\}=\mathrm{x}-\left[\mathrm{x}\right]\Rightarrow$$$$\mathrm{\Phi }={\int }_0^1\frac{\mathrm{dx}}{{\mathrm{e}}^{{(\mathrm{x}-\left[\mathrm{x}\right])}^2}+1}+{\int }_1^2\frac{\mathrm{dx}}{{\mathrm{e}}^{{(\mathrm{x}-\left[\mathrm{x}\right])}^2}+1}$$$$={\int }_0^1\frac{\mathrm{dx}}{1+{\mathrm{e}}^{{\mathrm{x}}^2}}+{\int }_1^2\frac{\mathrm{dx}}{1+{\mathrm{e}}^{{(\mathrm{x}-1)}^2}}(\to \mathrm{x}-1=\mathrm{t})$$$$={\int }_0^1\frac{\mathrm{dx}}{1+{\mathrm{e}}^{{\mathrm{x}}^2}}+{\int }_0^1\frac{\mathrm{dt}}{1+{\mathrm{e}}^{{\mathrm{t}}^2}}=2{\int }_0^1\frac{\mathrm{dx}}{1+{\mathrm{e}}^{{\mathrm{x}}^2}}$$$${\int }_0^1\frac{\mathrm{dx}}{1+{\mathrm{e}}^{{\mathrm{x}}^2}}={\int }_0^1\frac{{\mathrm{e}}^{-{\mathrm{x}}^2}}{1+{\mathrm{e}}^{-{\mathrm{x}}^2}}\mathrm{dx}={\int }_0^1{\mathrm{e}}^{-{\mathrm{x}}^2}\sum _{\mathrm{n}=0}^{\mathrm{\infty }}{\mathrm{e}}^{-{\mathrm{nx}}^2}\mathrm{dx}$$$$=\sum _{\mathrm{n}=0}^{\mathrm{\infty }}{\int }_0^1{\mathrm{e}}^{-(\mathrm{n}+1){\mathrm{x}}^2}\mathrm{dx}{=}_{\sqrt{\mathrm{n}+1}\mathrm{x}=\mathrm{z}}\sum _{\mathrm{n}=0}^{\mathrm{\infty }}{\int }_0^{\sqrt{\mathrm{n}+1}}{\mathrm{e}}^{-{\mathrm{z}}^2}\frac{\mathrm{dz}}{\mathrm{n}+1}$$$$\mathrm{\Phi }=\sum _{\mathrm{n}=0}^{\mathrm{\infty }}\frac{1}{\mathrm{n}+1}{\int }_0^{\sqrt{\mathrm{n}+1}}{\mathrm{e}}^{-{\mathrm{z}}^2}\mathrm{dz}$$

Commented by mathmax by abdo last updated on 29/Jun/21

$$\mathrm{\Phi }=\sum _{\mathrm{n}=0}^{\mathrm{\infty }}\frac{2}{\mathrm{n}+1}{\int }_0^{\sqrt{\mathrm{n}+1}}{\mathrm{e}}^{-{\mathrm{z}}^2}\mathrm{dz}$$

Commented by Dwaipayan Shikari last updated on 29/Jun/21

$$Thankssir$$

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