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Question Number 93227 by Ajao yinka last updated on 11/May/20

Commented by prakash jain last updated on 12/May/20

$$2{\int }_{-\mathrm{\infty }}^{+\mathrm{\infty }}{x}^4\delta (1+x+x^2+x^3)dx=2\times \left(\frac{2}{4}\right)=1$$$$\frac{1}{2+\frac{1}{2+\frac{1}{2+..}}}(\mathrm{it}\mathrm{is}\mathrm{covergent}.\mathrm{i}\mathrm{will}\mathrm{skip}\mathrm{proof})\mathrm{w}$$$$\frac{1}{2+x}=x\Rightarrow x^2+2x-1=0$$$$x=\frac{-2\pm \sqrt{8}}{2}=\sqrt{2}-1$$$$\mathrm{So}\mathrm{the}\mathrm{given}\mathrm{expression}\mathrm{becomes}$$$$1+\sqrt{2}-1+{\sinh }^{-1}\left(\underset{n=0}{\sum ^{\mathrm{\infty }}}\frac{1}{2n!(2n+1)}\right)-1=>$$$$=\sqrt{2}-1+{\sinh }^{-1}\left(\underset{n=0}{\sum ^{\mathrm{\infty }}}\frac{1}{2n!(2n+1)}\right)=0$$$$=\sqrt{2}-1+{\sinh }^{-1}\left(\sinh 1\right)$$$$=\sqrt{2}$$$$$$$$\mathrm{It}\mathrm{is}\mathrm{not}\mathrm{equal}\mathrm{to}\mathrm{zero}\mathrm{as}\mathrm{claimed}$$$$\mathrm{in}\mathrm{question}.\mathrm{Can}\mathrm{you}\mathrm{please}\mathrm{recheck}$$$$\mathrm{quesstiin}$$

Commented by prakash jain last updated on 12/May/20

For the proof of integral in last part. please see q93225

Commented by Ajao yinka last updated on 14/May/20

You're wrong

Commented by prakash jain last updated on 14/May/20

$$\mathrm{Do}\mathrm{you}\mathrm{also}\mathrm{want}\mathrm{go}\mathrm{explain}\mathrm{on}\mathrm{what}$$$$\mathrm{part}\mathrm{of}\mathrm{answer}\mathrm{is}\mathrm{wrong}?$$

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