Question Number 93227 by Ajao yinka last updated on 11/May/20
Commented by prakash jain last updated on 12/May/20
$$\mathrm{2}\int_{−\infty} ^{+\infty} {x}^{\mathrm{4}} \delta\left(\mathrm{1}+{x}+{x}^{\mathrm{2}} +{x}^{\mathrm{3}} \right){dx}=\mathrm{2}×\left(\frac{\mathrm{2}}{\mathrm{4}}\right)=\mathrm{1} \\ $$$$\frac{\mathrm{1}}{\mathrm{2}+\frac{\mathrm{1}}{\mathrm{2}+\frac{\mathrm{1}}{\mathrm{2}+..}}}\:\left(\mathrm{it}\:\mathrm{is}\:\mathrm{covergent}.\:\mathrm{i}\:\mathrm{will}\:\mathrm{skip}\:\mathrm{proof}\right)\mathrm{w} \\ $$$$\frac{\mathrm{1}}{\mathrm{2}+{x}}={x}\Rightarrow{x}^{\mathrm{2}} +\mathrm{2}{x}−\mathrm{1}=\mathrm{0} \\ $$$${x}=\frac{−\mathrm{2}\pm\sqrt{\mathrm{8}}}{\mathrm{2}}=\sqrt{\mathrm{2}}−\mathrm{1} \\ $$$$\mathrm{So}\:\mathrm{the}\:\mathrm{given}\:\mathrm{expression}\:\mathrm{becomes} \\ $$$$\mathrm{1}+\sqrt{\mathrm{2}}−\mathrm{1}+\mathrm{sinh}^{−\mathrm{1}} \left(\underset{{n}=\mathrm{0}} {\overset{\infty} {\sum}}\frac{\mathrm{1}}{\mathrm{2}{n}!\left(\mathrm{2}{n}+\mathrm{1}\right)}\right)−\mathrm{1}=> \\ $$$$=\sqrt{\mathrm{2}}−\mathrm{1}+\mathrm{sinh}^{−\mathrm{1}} \left(\underset{{n}=\mathrm{0}} {\overset{\infty} {\sum}}\frac{\mathrm{1}}{\mathrm{2}{n}!\left(\mathrm{2}{n}+\mathrm{1}\right)}\right)=\mathrm{0} \\ $$$$=\sqrt{\mathrm{2}}−\mathrm{1}+\mathrm{sinh}^{−\mathrm{1}} \left(\mathrm{sinh}\:\mathrm{1}\right) \\ $$$$=\sqrt{\mathrm{2}} \\ $$$$ \\ $$$$\mathrm{It}\:\mathrm{is}\:\mathrm{not}\:\mathrm{equal}\:\mathrm{to}\:\mathrm{zero}\:\mathrm{as}\:\mathrm{claimed} \\ $$$$\mathrm{in}\:\mathrm{question}.\:\mathrm{Can}\:\mathrm{you}\:\mathrm{please}\:\mathrm{recheck} \\ $$$$\mathrm{quesstiin} \\ $$
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
$$\mathrm{Do}\:\mathrm{you}\:\mathrm{also}\:\mathrm{want}\:\mathrm{go}\:\mathrm{explain}\:\mathrm{on}\:\mathrm{what} \\ $$$$\mathrm{part}\:\mathrm{of}\:\mathrm{answer}\:\mathrm{is}\:\mathrm{wrong}? \\ $$