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 $$ \\ $$$$\:\:\:\boldsymbol{{Calculate}}\:\boldsymbol{{the}}\:\boldsymbol{{following}}\:\boldsymbol{{integral}};\:\:\:\:\:\: \\ $$$$\:\:\int_{−\infty} ^{\infty} \int_{−\infty} ^{\infty} \int_{−\infty} ^{\infty} \boldsymbol{{xJ}}_{\mathrm{0}} \left(\sqrt{\boldsymbol{{x}}^{\mathrm{2}} +\boldsymbol{{y}}^{\mathrm{2}} }\right)\boldsymbol{{J}}_{\mathrm{1}} \left(\sqrt{\boldsymbol{{y}}^{\mathrm{2}} +\boldsymbol{{z}}^{\mathrm{2}}…
Question Number 218872 by Nicholas666 last updated on 16/Apr/25 $$ \\ $$$$\:\boldsymbol{{Calculate}}\:\boldsymbol{{the}}\:\boldsymbol{{following}}\:\boldsymbol{{integral}}; \\ $$$$\:\:\:\int_{\mathrm{0}} ^{\infty} \int_{\mathrm{0}} ^{\infty} \int_{\mathrm{0}} ^{\infty} \:\frac{\boldsymbol{{J}}_{\boldsymbol{\alpha}} \left(\boldsymbol{{ax}}\right)\boldsymbol{{J}}_{\boldsymbol{\beta}} \left(\boldsymbol{{by}}\right)\boldsymbol{{J}}_{\boldsymbol{\gamma}} \left(\boldsymbol{{cz}}\right)}{\:\sqrt{\boldsymbol{{x}}^{\mathrm{2}} +\boldsymbol{{y}}^{\mathrm{2}} +\boldsymbol{{z}}^{\mathrm{2}}…
Question Number 218866 by Nicholas666 last updated on 16/Apr/25 $$ \\ $$$$\:\:\:{evaluate}\:{the}\:{following}\:{integral}\:{in}\:{closed}\:{form}\:{or}\:{express} \\ $$$$\:{it}\:{in}\:{terms}\:{of}\:{known}\:{special}\:{functions};\:\:\: \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\int_{ } ^{\infty} \boldsymbol{{K}}_{\boldsymbol{{i}\lambda}} \left(\boldsymbol{{at}}\right)\boldsymbol{{J}}_{\boldsymbol{\nu}} \left(\boldsymbol{{bt}}\right)^{\boldsymbol{\mu}−\mathrm{1}} \boldsymbol{{dt}} \\ $$$$\:\boldsymbol{{where}}; \\…
Question Number 218748 by MrGaster last updated on 15/Apr/25 $$\int_{\mathrm{0}} ^{\infty} \frac{{x}^{\mathrm{2}} \mathrm{cos}\:{x}}{\mathrm{cosh}\:\mathrm{2}{x}−\mathrm{cos}\:{x}}−\frac{\mathrm{2}{x}^{\mathrm{2}} }{\mathrm{e}^{\mathrm{4}{x}} −\mathrm{2}{e}^{\mathrm{2}{x}} \mathrm{cos}\:{x}+\mathrm{1}}{dx},\mathrm{lemma}:\underset{{k}=\mathrm{1}} {\overset{+\infty} {\sum}}\frac{\mathrm{cos}\:{kx}}{{p}^{{k}} }=\frac{\mathrm{p}\:\mathrm{cos}\:{x}−\mathrm{1}}{{p}^{\mathrm{2}} −\mathrm{2}{p}\:\mathrm{cos}\:{x}+\mathrm{1}},{p}>\mathrm{1} \\ $$ Answered by MrGaster…
Question Number 218780 by Spillover last updated on 15/Apr/25 Commented by malwan last updated on 17/Apr/25 $${can}\:{you}\:{solve}\:{this}\:{integral} \\ $$$${please}\:? \\ $$ Answered by Spillover last…
Question Number 218781 by Ghisom last updated on 15/Apr/25 $$\mathrm{prove}: \\ $$$$\underset{\mathrm{0}} {\overset{\pi/\mathrm{4}} {\int}}\mathrm{arccos}\:\frac{\sqrt{\mathrm{2}}}{\:\sqrt{\mathrm{3}−\mathrm{tan}^{\mathrm{2}} \:{x}}}\:{dx}=\frac{\pi^{\mathrm{2}} }{\mathrm{24}} \\ $$ Answered by breniam last updated on 20/Apr/25…