Question Number 219078 by zetamaths last updated on 19/Apr/25 $$\left(\mathrm{1}−\frac{\mathrm{1}}{\mathrm{2}{m}}\right).\underset{{k}=\mathrm{1}} {\overset{{m}} {\sum}}\left(−\mathrm{1}\right)^{{k}−\mathrm{1}} \centerdot{k}\centerdot\frac{\left({m}!\right)^{\mathrm{2}} }{\left({m}−{k}\right)!\left({m}+{k}\right)!}=\frac{\mathrm{1}}{\mathrm{4}}\:\:\:\:\:\:{Proof}\:{this}\:{formula} \\ $$ Answered by MrGaster last updated on 19/Apr/25 $$\mathrm{Let}\:{S}\left({m}\right)\underset{{k}=\mathrm{1}} {\overset{{m}}…
Question Number 218907 by malwan last updated on 17/Apr/25 $$\:_{\mathrm{0}} \int^{\:\mathrm{45}} {arctan}\left(\frac{\mathrm{1}+{tan}\:{x}}{\:\sqrt{\mathrm{2}}}\right){dx}\:=\:? \\ $$ Commented by mr W last updated on 17/Apr/25 $${you}\:{should}\:{make}\:{clear}\:{what}\:{you} \\ $$$${mean}\:{with}\:\int_{\mathrm{0}}…
Question Number 218896 by Nicholas666 last updated on 17/Apr/25 $$ \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\boldsymbol{{prove}}; \\ $$$$\:\mid\int\int\int_{\left[\mathrm{0},\infty\right]^{\mathrm{3}} } \boldsymbol{{f}}\frac{\boldsymbol{{J}}_{\mathrm{0}} \left(\boldsymbol{{x}}\right)\boldsymbol{{J}}_{\mathrm{0}} \left(\boldsymbol{{y}}\right)\boldsymbol{{J}}_{\mathrm{0}} \left(\boldsymbol{{z}}\right)}{\mathrm{1}+\boldsymbol{{x}}^{\mathrm{2}} \boldsymbol{{y}}^{\mathrm{2}} \boldsymbol{{z}}^{\mathrm{2}} }\mid\leqslant\boldsymbol{{C}}\left(\int\int\int_{\mathbb{R}_{+} ^{\mathrm{3}} } \mid\boldsymbol{{f}}\mid\left(\mathrm{1}+\boldsymbol{{x}}^{\mathrm{2}}…
Question Number 218952 by Spillover last updated on 17/Apr/25 Commented by MathematicalUser2357 last updated on 17/Apr/25 $$\mathrm{What}\:\mathrm{kind}\:\mathrm{of}\:\mathrm{function}\:\mathrm{is}\:\mathrm{Ti}_{\mathrm{2}} ???\:\mathrm{Can}\:\mathrm{someone}\:\mathrm{help}\:\mathrm{me}\:\mathrm{before}\:\mathrm{I}\:\mathrm{eat}\:\mathrm{that}\:\mathrm{question}??? \\ $$ Commented by SdC355 last updated…
Question Number 218949 by Spillover last updated on 17/Apr/25 Commented by Spillover last updated on 17/Apr/25 $${ans}=\frac{\mathrm{ln}\:\left(\mathrm{1}+\sqrt{\left.\mathrm{2}\right)}\right.}{\:\sqrt{\mathrm{2}}} \\ $$ Commented by Nicholas666 last updated on…
Question Number 218950 by Spillover last updated on 17/Apr/25 Commented by Spillover last updated on 17/Apr/25 $${ans}=\frac{\mathrm{2}^{\mathrm{2}{n}} }{\left(\mathrm{2}{n}\right)!}\pi \\ $$ Answered by Spillover last updated…
Question Number 218951 by Spillover last updated on 17/Apr/25 Commented by Spillover last updated on 17/Apr/25 $${ans}=\frac{{z}}{\mathrm{sinh}\:{z}} \\ $$ Answered by Spillover last updated on…
Question Number 218879 by Nicholas666 last updated on 16/Apr/25 $$ \\ $$$$\:\:\:\boldsymbol{{Calculate}}\:\boldsymbol{{the}}\:\boldsymbol{{following}}\:\boldsymbol{{integral}};\:\:\:\:\:\: \\ $$$$\:\:\int_{−\infty} ^{\infty} \int_{−\infty} ^{\infty} \int_{−\infty} ^{\infty} \boldsymbol{{xJ}}_{\mathrm{0}} \left(\sqrt{\boldsymbol{{x}}^{\mathrm{2}} +\boldsymbol{{y}}^{\mathrm{2}} }\right)\boldsymbol{{J}}_{\mathrm{1}} \left(\sqrt{\boldsymbol{{y}}^{\mathrm{2}} +\boldsymbol{{z}}^{\mathrm{2}}…
Question Number 218872 by Nicholas666 last updated on 16/Apr/25 $$ \\ $$$$\:\boldsymbol{{Calculate}}\:\boldsymbol{{the}}\:\boldsymbol{{following}}\:\boldsymbol{{integral}}; \\ $$$$\:\:\:\int_{\mathrm{0}} ^{\infty} \int_{\mathrm{0}} ^{\infty} \int_{\mathrm{0}} ^{\infty} \:\frac{\boldsymbol{{J}}_{\boldsymbol{\alpha}} \left(\boldsymbol{{ax}}\right)\boldsymbol{{J}}_{\boldsymbol{\beta}} \left(\boldsymbol{{by}}\right)\boldsymbol{{J}}_{\boldsymbol{\gamma}} \left(\boldsymbol{{cz}}\right)}{\:\sqrt{\boldsymbol{{x}}^{\mathrm{2}} +\boldsymbol{{y}}^{\mathrm{2}} +\boldsymbol{{z}}^{\mathrm{2}}…
Question Number 218866 by Nicholas666 last updated on 16/Apr/25 $$ \\ $$$$\:\:\:{evaluate}\:{the}\:{following}\:{integral}\:{in}\:{closed}\:{form}\:{or}\:{express} \\ $$$$\:{it}\:{in}\:{terms}\:{of}\:{known}\:{special}\:{functions};\:\:\: \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\int_{ } ^{\infty} \boldsymbol{{K}}_{\boldsymbol{{i}\lambda}} \left(\boldsymbol{{at}}\right)\boldsymbol{{J}}_{\boldsymbol{\nu}} \left(\boldsymbol{{bt}}\right)^{\boldsymbol{\mu}−\mathrm{1}} \boldsymbol{{dt}} \\ $$$$\:\boldsymbol{{where}}; \\…