Question Number 69236 by ~ À ® @ 237 ~ last updated on 21/Sep/19 $${Use}\:\:{Residus}\:{theorem}\:{to}\:{prove}\:{that}\:\forall\:{a}>\mathrm{0}\:\:\underset{{n}=\mathrm{0}} {\overset{\infty} {\sum}}\:\frac{\mathrm{1}}{\:{n}^{\mathrm{2}} +{a}^{\mathrm{2}} }\:=\:\frac{\mathrm{1}}{\mathrm{2}}\left(\frac{\pi}{{ash}\left(\pi{a}\right)}\:\:\:−\frac{\mathrm{1}}{{a}^{\mathrm{2}} }\right) \\ $$$${and}\:\:\:\:\:\:\underset{{n}=\mathrm{0}} {\overset{\infty} {\sum}}\:\frac{\left(−\mathrm{1}\right)^{{n}} }{{n}^{\mathrm{2}}…
Question Number 134769 by abdurehime last updated on 07/Mar/21 Commented by abdurehime last updated on 07/Mar/21 $$\mathrm{please}\:\mathrm{be}?\mathrm{fast} \\ $$ Commented by john_santu last updated on…
Question Number 134768 by abdurehime last updated on 07/Mar/21 Commented by abdurehime last updated on 07/Mar/21 $$\mathrm{be}\:\mathrm{fast}\:\mathrm{please}\:\mathrm{help}\:\mathrm{me}\:\mathrm{with}\:\mathrm{in}\:\mathrm{10}\:\mathrm{min} \\ $$ Commented by mr W last updated…
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Question Number 69233 by ~ À ® @ 237 ~ last updated on 21/Sep/19 $${Prove}\:{that}\:\:{B}=\int_{\mathrm{0}} ^{\mathrm{1}} \:\:\:\left[{ln}\left(−{lnu}\right)\right]^{\mathrm{2}} \:{du}\:=\:\gamma^{\mathrm{2}} +\:\zeta\left(\mathrm{2}\right)\:\: \\ $$ Commented by mathmax by…
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Question Number 3695 by Rasheed Soomro last updated on 19/Dec/15 $$\mathcal{C}{an}\:{we}\:{say}\:{that} \\ $$$$\mathcal{A}\:{line}\:{is}\:{a}\:{circle}\:{whose}\:{radius}\:{is}\:\infty \\ $$$$\mathcal{O}{r} \\ $$$${A}\:{circle}\:{with}\:\infty\:{radius}\:{is}\:{a}\:{line}\:\:? \\ $$ Commented by Filup last updated on…
Question Number 134764 by bramlexs22 last updated on 07/Mar/21 $$\:\mathbb{Z}\:=\:\int_{\mathrm{0}} ^{\:\mathrm{2}} \:\frac{\mathrm{dx}}{\:\sqrt{\mid\mathrm{x}−\mathrm{1}\mid}}\:? \\ $$ Answered by EDWIN88 last updated on 07/Mar/21 $$\mathbb{Z}\:=\:\int_{\mathrm{0}} ^{\:\mathrm{2}} \:\frac{\mathrm{dx}}{\:\sqrt{\mid\mathrm{x}−\mathrm{1}\mid}}\:;\:\mathrm{let}\:\mathrm{u}=\mid\mathrm{x}−\mathrm{1}\mid \\…
Question Number 69231 by ~ À ® @ 237 ~ last updated on 21/Sep/19 $${Prove}\:{that}\:\:\underset{{p}=\mathrm{0}} {\overset{\infty} {\sum}}\:\:\frac{\left(−\mathrm{1}\right)^{{p}} }{\mathrm{4}{p}+\mathrm{1}}\:=\:\frac{\pi−{argcoth}\left(\sqrt{\mathrm{2}}\:\right)}{\mathrm{4}\sqrt{\mathrm{2}}}\:\:{and} \\ $$$$\underset{{p}=\mathrm{0}} {\overset{\infty\:} {\sum}}\:\frac{\left(−\mathrm{1}\right)^{{p}} }{\mathrm{4}{p}+\mathrm{3}}\:=\:\frac{\pi+{argcoth}\left(\sqrt{\mathrm{2}}\:\right)}{\mathrm{4}\sqrt{\mathrm{2}}\:}\:\:\: \\ $$…
Question Number 134767 by jagadharkumar last updated on 07/Mar/21 Terms of Service Privacy Policy Contact: info@tinkutara.com