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Question Number 142528 by mohammad17 last updated on 01/Jun/21

$${\int }_1^{\mathrm{\infty }}\frac{dx}{e^x-2^x}$$

Answered by mathmax by abdo last updated on 02/Jun/21

$$\mathrm{\Phi }={\int }_1^{\mathrm{\infty }}\frac{\mathrm{dx}}{{\mathrm{e}}^{\mathrm{x}}-2^{\mathrm{x}}}\Rightarrow \mathrm{\Phi }={\int }_1^{\mathrm{\infty }}\frac{{\mathrm{e}}^{-\mathrm{x}}}{1-2^{\mathrm{x}}{\mathrm{e}}^{-\mathrm{x}}}\mathrm{dx}={\int }_1^{\mathrm{\infty }}\frac{{\mathrm{e}}^{-\mathrm{x}}\mathrm{dx}}{1-{\left(\frac{2}{\mathrm{e}}\right)}^{\mathrm{x}}}(\mid \frac{2}{\mathrm{e}}\mid <1)$$$$={\int }_1^{\mathrm{\infty }}{\mathrm{e}}^{-\mathrm{x}}\sum _{\mathrm{n}=0}^{\mathrm{\infty }}{\left(\frac{2}{\mathrm{e}}\right)}^{\mathrm{nx}}\mathrm{dx}=\sum _{\mathrm{n}=0}^{\mathrm{\infty }}{\int }_1^{\mathrm{\infty }}{\mathrm{e}}^{-\mathrm{x}}{2}^{\mathrm{nx}}{\mathrm{e}}^{-\mathrm{nx}}\mathrm{dx}$$$$=\sum _{\mathrm{n}=0}^{\mathrm{\infty }}{\int }_1^{\mathrm{\infty }}{\mathrm{e}}^{-(\mathrm{n}+1)\mathrm{x}}{2}^{\mathrm{nx}}\mathrm{dx}=\sum _{\mathrm{n}=0}^{\mathrm{\infty }}{\int }_1^{\mathrm{\infty }}{\mathrm{e}}^{-(\mathrm{n}+1)\mathrm{x}}{\mathrm{e}}^{\mathrm{nxlog}2}\mathrm{ex}$$$$=\sum _{\mathrm{n}=0}^{\mathrm{\infty }}{\int }_1^{\mathrm{\infty }}{\mathrm{e}}^{(\mathrm{nlog}2-\mathrm{n}-1)\mathrm{x}}\mathrm{dx}=\sum _{\mathrm{n}=0}^{\mathrm{\infty }}{\left[\frac{1}{\mathrm{nlog}2-\mathrm{n}-1}{\mathrm{e}}^{(\mathrm{nlog}2-\mathrm{n}-1)\mathrm{x}}\right]}_1^{\mathrm{\infty }}$$$$=-\sum _{\mathrm{n}=0}^{\mathrm{\infty }}\frac{1}{\mathrm{nlog}2-\mathrm{n}-1}=\sum _{\mathrm{n}=0}^{\mathrm{\infty }}\frac{1}{\mathrm{n}+1-\mathrm{nlog}2}$$$$\Rightarrow \mathrm{\Phi }=1+\frac{1}{2-\log 2}+\frac{1}{3-2\log 2}+\frac{1}{4-3\log 2}+....$$

Commented by mohammad17 last updated on 02/Jun/21

$$thankyousircqnyoucomplete$$

Commented by Mathspace last updated on 02/Jun/21

$$thisistheanswerbyseries...$$

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