Author: Tinku Tara

Question Number 219359 by SdC355 last updated on 23/Apr/25 $$\mathrm{is}\:\underset{{k}=−\infty} {\overset{\infty} {\sum}}\:{J}_{\nu} \left({k}\right)= \\ $$$$\underset{{k}=−\infty} {\overset{\infty} {\sum}}\:\underset{{l}=\mathrm{0}} {\overset{\infty} {\sum}}\:\frac{\left(−\mathrm{1}\right)^{{l}} }{{l}!\left({l}+\nu\right)!}\left(\frac{{k}}{\mathrm{2}}\right)^{\mathrm{2}{l}+\nu} =\underset{{l}=\mathrm{0}} {\overset{\infty} {\sum}}\:\underset{{k}=−\infty} {\overset{\infty} {\sum}}\:\frac{\left(−\mathrm{1}\right)^{{l}}…

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 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) \\…