Question Number 69257 by Henri Boucatchou last updated on 21/Sep/19 $$\int_{−\mathrm{2}} ^{\:\:\mathrm{4}} \mid\boldsymbol{{x}}\mid^{\mathrm{2}\boldsymbol{{x}}^{\mathrm{3}} } \boldsymbol{{dx}}\:=\:? \\ $$ Terms of Service Privacy Policy Contact: info@tinkutara.com
Question Number 69246 by ~ À ® @ 237 ~ last updated on 21/Sep/19 $$\:\:\:{Explicit}\:\:\int_{\mathrm{0}} ^{\infty} \:\:\frac{{Si}\left({ax}\right)}{{x}+{b}}\:{dx}\:\:\:{with}\:\:{Si}\left({u}\right)=\int_{\mathrm{0}} ^{{u}} \:\frac{{sinx}}{{x}}{dx} \\ $$ Terms of Service Privacy…
Question Number 69241 by ~ À ® @ 237 ~ last updated on 21/Sep/19 $$\:{Let}\:{consider}\:{K}=\int_{\mathrm{0}} ^{\mathrm{1}} \:\frac{\left(\mathrm{1}−{x}^{{a}} \right)\left(\mathrm{1}−{x}^{{b}} \right)\left(\mathrm{1}−{x}^{{c}} \right)}{\left({x}−\mathrm{1}\right){lnx}}{dx}\: \\ $$$${prove}\:{that}\: \\ $$$${e}^{{K}} =\:\frac{\left({a}+{b}\right)!\left({a}+{c}\right)!\left({b}+{c}\right)!}{{a}!{b}!{c}!\left({a}+{b}+{c}\right)!}\:\:…
Question Number 69238 by ~ À ® @ 237 ~ last updated on 21/Sep/19 $${Use}\:{Residus}\:{Theorem}\:{to}\:{explicit}\: \\ $$$${f}\left({a}\right)=\underset{{n}=\mathrm{1}} {\overset{\infty} {\sum}}\:\frac{\left(−\mathrm{1}\right)^{{n}} {sin}\left({na}\right)}{{n}^{\mathrm{3}} }\:\: \\ $$ Commented by…
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…
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 134750 by mnjuly1970 last updated on 06/Mar/21 $$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:…..{advanced}\:\:\:{calculus}…. \\ $$$$\:\:\:{prove}\:\:\:{that}::\: \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\boldsymbol{\phi}=\Sigma\frac{\mathrm{1}}{{n}^{\mathrm{3}} {sin}\left(\sqrt{\mathrm{2}}\:\pi{n}\right)}\:=\frac{−\mathrm{13}\sqrt{\mathrm{2}}\:\pi^{\mathrm{3}} \:}{\mathrm{720}} \\ $$$$\:\:\:\:\:…{m}.{n}… \\ $$ Terms of Service Privacy Policy…
Question Number 134716 by metamorfose last updated on 06/Mar/21 $$\int\frac{{ln}\left({x}\right)}{{x}−\mathrm{1}}{dx}=…?? \\ $$ Answered by Lordose last updated on 06/Mar/21 $$\int\frac{\mathrm{ln}\left(\mathrm{x}\right)}{\mathrm{x}−\mathrm{1}}\mathrm{dx}\:\overset{\mathrm{u}=\mathrm{x}−\mathrm{1}} {=}\int\frac{\mathrm{ln}\left(\mathrm{1}+\mathrm{u}\right)}{\mathrm{u}}\mathrm{du}\:=\:−\mathrm{Li}_{\mathrm{2}} \left(−\mathrm{u}\right)\:+\:\mathrm{C} \\ $$$$\Omega\:=\:−\mathrm{Li}_{\mathrm{2}} \left(\mathrm{1}−\mathrm{x}\right)\:+\:\mathrm{C}…
Question Number 134708 by bramlexs22 last updated on 06/Mar/21 $$\mathscr{D}\:=\:\int_{\mathrm{0}} ^{\:\pi/\mathrm{2}} \mathrm{sin}\:^{\mathrm{4}} \mathrm{x}\:\mathrm{cos}\:^{\mathrm{5}} \mathrm{x}\:\mathrm{dx}\: \\ $$ Answered by john_santu last updated on 06/Mar/21 $$\mathscr{D}\:=\:\int_{\mathrm{0}} ^{\:\pi/\mathrm{2}}…