Question Number 219355 by SdC355 last updated on 23/Apr/25 $$\int_{\mathrm{0}} ^{\:\infty} \:\:\frac{\mathrm{sin}\left({z}\right)}{{z}}{e}^{−{zt}} \mathrm{d}{z}=?? \\ $$ Answered by breniam last updated on 23/Apr/25 $$={I}\left({t}\right) \\ $$$${I}'\left({t}\right)=\underset{{t}_{\mathrm{0}}…
Question Number 219354 by SdC355 last updated on 23/Apr/25 $$\int_{−\infty} ^{\:+\infty} \:\:{ze}^{−{z}^{\mathrm{3}} } \:\mathrm{d}{z}=?? \\ $$ Answered by breniam last updated on 24/Apr/25 $$\underset{−\infty} {\overset{\mathrm{0}}…
Question Number 219349 by SdC355 last updated on 23/Apr/25 $$\int_{\mathrm{0}} ^{\:\infty} \:{e}^{−{r}^{\mathrm{2}} } \mathrm{cos}\left({r}\right)\:\mathrm{d}{r}=?? \\ $$ Answered by vnm last updated on 23/Apr/25 $$\int_{\mathrm{0}} ^{\infty}…
Question Number 219348 by SdC355 last updated on 23/Apr/25 $$\int_{\mathrm{0}} ^{\:\infty} \:\:{Y}_{\mathrm{0}} \left({z}\right){e}^{−\mathrm{2}{z}} \mathrm{d}{z}=?? \\ $$$${Y}_{\nu} \left({z}\right)\:\mathrm{is}\:\mathrm{Second}\:\mathrm{Bessel}\:\mathrm{Function} \\ $$ Terms of Service Privacy Policy Contact:…
Question Number 219350 by SdC355 last updated on 23/Apr/25 $$\int\:\:\frac{\mathrm{1}}{\mathrm{cos}\left({u}\right)+\mathrm{sin}\left({u}\right)+\mathrm{1}}\:\mathrm{d}{u}=?? \\ $$ Answered by vnm last updated on 23/Apr/25 $$\int\frac{\mathrm{1}}{\mathrm{2cos}^{\mathrm{2}} \frac{{u}}{\mathrm{2}}+\mathrm{2sin}\frac{{u}}{\mathrm{2}}\mathrm{cos}\frac{{u}}{\mathrm{2}}}\mathrm{d}{u}= \\ $$$$\int\frac{\mathrm{1}}{\left(\mathrm{1}+\mathrm{tan}\frac{{u}}{\mathrm{2}}\right)\centerdot\mathrm{2cos}^{\mathrm{2}} \frac{{u}}{\mathrm{2}}}\mathrm{d}{u}= \\…
Question Number 219344 by SdC355 last updated on 23/Apr/25 $$\int_{\mathrm{0}} ^{\:\infty} \:\frac{\mathrm{cos}\left({z}\right)}{\:\sqrt{{z}^{\mathrm{2}} +\mathrm{1}}}\:\mathrm{d}{z}=?? \\ $$ Answered by Nicholas666 last updated on 23/Apr/25 $${K}_{\mathrm{0}} \left(\mathrm{1}\right) \\…
Question Number 219345 by SdC355 last updated on 23/Apr/25 $$\int_{−\infty} ^{+\infty} \int_{−\infty} ^{\:+\infty} \:\:−\frac{{x}}{\:\sqrt{{x}^{\mathrm{2}} +{y}^{\mathrm{2}} }}\:\mathrm{da}=???\:\: \\ $$ Terms of Service Privacy Policy Contact: info@tinkutara.com
Question Number 219347 by SdC355 last updated on 23/Apr/25 $$\int_{\mathrm{0}} ^{\:\infty} \:{K}_{\mathrm{0}} \left({z}\right){e}^{−{kz}} \mathrm{d}{z}=??? \\ $$$${K}_{\nu} \left({z}\right)\:\mathrm{is}\:\mathrm{modified}\:\mathrm{Bessel}\:\mathrm{function} \\ $$ Answered by Nicholas666 last updated on…
Question Number 219346 by SdC355 last updated on 23/Apr/25 $$\int_{\mathrm{0}} ^{\:\infty} \mathrm{ln}\left({z}\right){e}^{−\mathrm{3}{z}} \mathrm{d}{z}=??? \\ $$ Terms of Service Privacy Policy Contact: info@tinkutara.com
Question Number 219340 by SdC355 last updated on 23/Apr/25 $$\overset{\rightarrow} {\boldsymbol{\mathrm{F}}}\left({x},{y},{z}\right)=−\frac{{x}}{\:\sqrt{{x}^{\mathrm{2}} +{y}^{\mathrm{2}} +{z}^{\mathrm{2}} }}\overset{\rightarrow} {\boldsymbol{\mathrm{e}}}_{\mathrm{1}} −\frac{{y}}{\:\sqrt{{x}^{\mathrm{2}} +{y}^{\mathrm{2}} +{z}^{\mathrm{2}} }}\overset{\rightarrow} {\boldsymbol{\mathrm{e}}}_{\mathrm{2}} −\frac{{z}}{\:\sqrt{{x}^{\mathrm{2}} +{y}^{\mathrm{2}} +{z}^{\mathrm{2}} }}\overset{\rightarrow} {\boldsymbol{\mathrm{e}}}_{\mathrm{3}}…