Question Number 219625 by Nicholas666 last updated on 29/Apr/25 $$ \\ $$$$\:\mathrm{Determine}\:\mathrm{all}\:\mathrm{real}\:\mathrm{numbers}\:{x}\: \\ $$$$\:\:\:\mathrm{that}\:\mathrm{statisfy}\:\mathrm{the}\:\mathrm{following}\:\mathrm{inequality}; \\ $$$$\:\:\sqrt{{x}^{\mathrm{4}} +\mathrm{1}}+\sqrt{{x}^{\mathrm{4}} +\mathrm{4}{x}^{\mathrm{2}} +\mathrm{4}}\:\leqslant\:\sqrt{\left({x}^{\mathrm{2}} +\mathrm{1}\right)\left({x}^{\mathrm{2}} +\mathrm{4}\right)}\:+\:\mid{x}\mid\sqrt{{x}^{\mathrm{2}} +\mathrm{3}}\:\:\:\:\:\: \\ $$$$ \\…
Question Number 219617 by Ghisom last updated on 29/Apr/25 $$\mathrm{ok},\:\mathrm{let}'\mathrm{s}\:\mathrm{all}\:\mathrm{answer}\:\mathrm{questions}\:\mathrm{from}\:\mathrm{anywhere} \\ $$$$\mathrm{on}\:\mathrm{the}\:{www}\:\mathrm{using}\:\mathrm{the}\:\mathrm{given}\:\mathrm{results}\:\mathrm{from} \\ $$$$\mathrm{the}\:\mathrm{sources}\:\mathrm{or}\:{wolframalpha}\:\mathrm{or}\:\mathrm{any}\:\mathrm{AI} \\ $$$$\mathrm{available}.\:\mathrm{this}\:\mathrm{promises}\:\mathrm{great}\:\mathrm{fun}! \\ $$ Terms of Service Privacy Policy Contact: info@tinkutara.com
Question Number 219602 by SdC355 last updated on 29/Apr/25 $$\int_{\mathrm{0}} ^{\:\infty} \:\mathrm{d}{t}\:{e}^{−\boldsymbol{{i}}{kt}} {J}_{−\frac{\mathrm{2}}{\mathrm{3}}} \left({t}\right)−\int_{\mathrm{0}} ^{\:\infty} \:\mathrm{d}{t}\:{e}^{−\boldsymbol{{i}}{kt}} {Y}_{−\frac{\mathrm{2}}{\mathrm{3}}} \left({t}\right)=?? \\ $$ Commented by Nicholas666 last updated…
Question Number 219597 by SdC355 last updated on 29/Apr/25 $$\mathrm{Evaluate}\:\mathrm{integral}\:\mathrm{by}\:\mathrm{Complex}\:\mathrm{integral}\:\mathrm{method} \\ $$$$\int_{\mathrm{0}} ^{\:\mathrm{2}\pi} \:\:\frac{\mathrm{1}}{{a}+{b}\centerdot\mathrm{cos}\left({n}\theta\right)}\:\mathrm{d}\theta \\ $$ Answered by Nicholas666 last updated on 29/Apr/25 $$\frac{\mathrm{2}\pi}{{n}\sqrt{{a}^{\mathrm{2}} −{b}^{\mathrm{2}}…
Question Number 219591 by SdC355 last updated on 29/Apr/25 $$\boldsymbol{\mathrm{LT}}\left\{\frac{\mathrm{Ai}^{\left(\mathrm{1}\right)} \left(−\left(\frac{\mathrm{3}}{\mathrm{2}}\right)^{\mathrm{2}/\mathrm{3}} {z}^{\mathrm{2}/\mathrm{3}} \right)+\sqrt{\mathrm{3}}\mathrm{Bi}^{\left(\mathrm{1}\right)} \left(−\left(\frac{\mathrm{3}}{\mathrm{2}}\right)^{\mathrm{2}/\mathrm{3}} {z}^{\mathrm{2}/\mathrm{3}} \right.}{\:^{\mathrm{3}} \sqrt{\mathrm{2}}\centerdot^{\mathrm{6}} \sqrt{\mathrm{3}}{z}^{\mathrm{2}/\mathrm{3}} }\right\}=??? \\ $$$$\boldsymbol{\mathrm{LT}}\left\{\ast\right\}=\int_{\mathrm{0}} ^{\:\infty} \:{e}^{−{zt}} \ast \\…
Question Number 219496 by SdC355 last updated on 27/Apr/25 $${p}\left({t}\right)=−\frac{\mathrm{1}}{\boldsymbol{{i}}\pi}\int_{−\infty\boldsymbol{{i}}+\boldsymbol{\gamma}} ^{\:\infty\boldsymbol{{i}}+\boldsymbol{\gamma}} \:\:\frac{{e}^{{st}} \left(\mathrm{ln}\left({s}\right)+\boldsymbol{\gamma}_{\mathrm{0}} \right)}{{s}}\:\mathrm{d}{s} \\ $$$${q}\left({t}\right)=\frac{\mathrm{1}}{\boldsymbol{{i}}\pi}\int_{−\infty\boldsymbol{{i}}+\boldsymbol{\gamma}} ^{\:\:\infty\boldsymbol{{i}}+\boldsymbol{\gamma}} \:\left\{−\frac{\pi}{\mathrm{2}{s}}\boldsymbol{\mathrm{L}}_{\mathrm{0}} \left({s}\right)+\frac{\pi}{\mathrm{2}{s}}\boldsymbol{{i}}{Y}_{\mathrm{0}} \left(−\boldsymbol{{i}}{s}\right)\right\}{e}^{{st}} \:\mathrm{d}{s} \\ $$$$\mathrm{g}\left({s}\right)=\int_{\mathrm{0}} ^{\:\infty} \:\:{J}_{\nu}…
Question Number 219481 by SdC355 last updated on 26/Apr/25 $$\mathrm{pls}\:\mathrm{Help}…. \\ $$$$\mathrm{1}.\:−\frac{\mathrm{1}}{\boldsymbol{{i}}\pi}\:\int_{\:−\infty+\boldsymbol{{i}\zeta}_{\mathrm{0}} } ^{\:\infty+\boldsymbol{{i}\zeta}_{\mathrm{0}} } \:\frac{{e}^{{st}} \left(\mathrm{ln}\left({t}\right)+\boldsymbol{\gamma}\right)}{{t}}\:\mathrm{d}{t}\: \\ $$$$\mathrm{2}.\:\int_{\mathrm{0}} ^{\:\infty} \:\:{t}\centerdot{J}_{\nu} \left({t}\right){J}_{\nu} \left({rt}\right)\:\mathrm{d}{t} \\ $$$$\mathrm{3}.\:\int_{\mathrm{0}}…
Question Number 219465 by SdC355 last updated on 26/Apr/25 $${prove} \\ $$$$\int_{\mathrm{0}} ^{\:\infty} \:\:\sqrt{{r}^{\mathrm{2}} −{t}^{\mathrm{2}} }{e}^{−{pt}} \mathrm{d}{t}=\frac{−{r}\pi\boldsymbol{\mathrm{L}}_{\mathrm{1}} \left({rp}\right)+\pi{rI}_{\mathrm{1}} \left({up}\right)+\mathrm{2}\boldsymbol{{i}}{rK}_{\mathrm{1}} \left({rp}\right)}{\mathrm{2}{p}} \\ $$$$\boldsymbol{\mathrm{L}}_{\nu} \left({x}\right)\:\mathrm{is}\:\mathrm{Modified}\:\mathrm{Struve}\:\mathrm{function} \\ $$$${I}_{\nu}…
Question Number 219445 by SdC355 last updated on 25/Apr/25 $${prove} \\ $$$$\int_{\mathrm{0}} ^{\:\infty} \:\sqrt{{r}^{\mathrm{2}} −{t}^{\mathrm{2}} }{e}^{−{wt}} \mathrm{d}{t}=\frac{−{r}\pi\boldsymbol{\mathrm{L}}_{\mathrm{1}} \left({rw}\right)+{r}\pi{I}_{\mathrm{1}} \left({rw}\right)+\mathrm{2}{r}\boldsymbol{{i}}{K}_{\mathrm{1}} \left({rw}\right)}{\mathrm{2}{w}} \\ $$ Commented by MrGaster…
Question Number 219423 by SdC355 last updated on 24/Apr/25 Answered by SdC355 last updated on 24/Apr/25 $$\overset{\rightarrow} {\boldsymbol{\mathcal{S}}}\left({u},{v}\right)=\begin{cases}{\frac{\mathrm{5}}{\mathrm{4}}\left(\mathrm{1}−\frac{{v}}{\mathrm{2}\pi}\right)\mathrm{cos}\left(\mathrm{2}{v}\right)\left(\mathrm{1}+\mathrm{cos}\left({u}\right)\right)+\mathrm{cos}\left(\mathrm{2}{v}\right)}\\{\frac{\mathrm{5}}{\mathrm{4}}\left(\mathrm{1}−\frac{{v}}{\mathrm{2}\pi}\right)\mathrm{sin}\left(\mathrm{2}{v}\right)\left(\mathrm{1}+\mathrm{cos}\left({u}\right)\right)+\mathrm{sin}\left(\mathrm{2}{v}\right)}\\{\frac{\mathrm{10}{v}}{\mathrm{2}\pi}+\frac{\mathrm{5}}{\mathrm{4}}\left(\mathrm{1}−\frac{{v}}{\mathrm{2}\pi}\right)\mathrm{sin}\left({u}\right)}\end{cases} \\ $$$$\mathrm{0}\leq{u}\leq\mathrm{2}\pi\:,\:−\pi\leq{v}\leq\mathrm{2}\pi \\ $$$$\overset{\rightarrow} {\boldsymbol{\mathrm{F}}}\left({x},{y},{z}\right)=−{x}\overset{\rightarrow} {\boldsymbol{\mathrm{e}}}_{\mathrm{1}} −{y}\overset{\rightarrow}…