Question Number 33987 by abdo imad last updated on 28/Apr/18 $${find}\:\int_{−\infty} ^{+\infty} \:\:\:\frac{{cos}\left(\alpha{x}\right)}{\left(\mathrm{1}+{x}^{\mathrm{2}} \right)^{\mathrm{3}} }\:{dx}\:{with}\:\alpha\geqslant\mathrm{0}\:. \\ $$ Commented by abdo mathsup 649 cc last updated…
Question Number 33986 by abdo imad last updated on 28/Apr/18 $${find}\:\int_{−\infty} ^{+\infty} \:\:\frac{{cos}\left({tx}\right)}{\left(\mathrm{1}+{x}^{\mathrm{2}} \right)^{\mathrm{2}} }\:{dx}\:{with}\:{t}\geqslant\mathrm{0} \\ $$ Commented by math khazana by abdo last updated…
Question Number 33983 by abdo imad last updated on 28/Apr/18 $${find}\:\int_{\mathrm{1}\left(\right.} ^{\infty} \frac{\mathrm{1}}{{x}}{ln}\left(\frac{{x}+\mathrm{1}}{{x}−\mathrm{1}}\right){dx}. \\ $$ Commented by abdo mathsup 649 cc last updated on 03/May/18…
Question Number 33984 by abdo imad last updated on 28/Apr/18 $${calculate}\:\int_{\mathrm{0}} ^{\mathrm{1}} \:\:\frac{\mathrm{1}}{{x}}{ln}\left(\frac{\mathrm{1}+{x}}{\mathrm{1}−{x}}\right){dx} \\ $$ Commented by prof Abdo imad last updated on 29/Apr/18 $${let}\:{put}\:\:{I}\:=\:\int_{\mathrm{0}}…
Question Number 99516 by 175 last updated on 21/Jun/20 Answered by mathmax by abdo last updated on 21/Jun/20 $$\mathrm{I}\:=\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{4}}} \:\frac{\mathrm{dx}}{\mathrm{cos}^{\mathrm{4}} \mathrm{x}+\mathrm{sin}^{\mathrm{4}} \mathrm{x}−\mathrm{cos}^{\mathrm{2}} \mathrm{xsin}^{\mathrm{2}} \mathrm{x}}\:\Rightarrow\:\mathrm{I}\:=\int_{\mathrm{0}}…
Question Number 33980 by abdo imad last updated on 28/Apr/18 $$\left.\mathrm{1}\right)\:{let}\:{consider}\:{f}\left({x}\right)=\mid{cosx}\mid\:\pi\:{periodix} \\ $$$${developp}\:{f}\:{at}\:{fourier}\:{serie} \\ $$$$\left.\mathrm{2}\right){find}\:{the}\:{valueof}\:\sum_{{n}=\mathrm{1}} ^{\infty} \:\:\frac{\left(−\mathrm{1}\right)^{{n}} }{\mathrm{4}{n}^{\mathrm{2}} \:−\mathrm{1}} \\ $$$$\left.\mathrm{3}\right){find}\:{the}\:{value}\:{of}\:\:\sum_{{n}=\mathrm{1}} ^{\infty} \:\:\:\frac{\mathrm{1}}{\left(\mathrm{4}{n}^{\mathrm{2}} −\mathrm{1}\right)^{\mathrm{2}} }\:.…
Question Number 33978 by abdo imad last updated on 28/Apr/18 $${let}\:{f}\left({t}\right)\:=\:\int_{\mathrm{0}} ^{\infty} \:\:\:\frac{{sin}\left({x}^{\mathrm{2}} \right){e}^{−{tx}^{\mathrm{2}} } }{{x}^{\mathrm{2}} }\:{dx}\:\:\:\:\:\:\:{with}\:{t}>\mathrm{0} \\ $$$${find}\:{a}\:{simple}\:{form}\:{of}\:{f}^{'} \left({t}\right)\:. \\ $$ Commented by math…
Question Number 33979 by abdo imad last updated on 28/Apr/18 $${we}\:{give}\:{for}\:{t}>\mathrm{0}\:\int_{\mathrm{0}} ^{\infty} \:\:\frac{{sinx}}{{x}}\:{e}^{−{tx}} {dx}\:=\frac{\pi}{\mathrm{2}}\:−{arctant} \\ $$$${use}\:{this}\:{result}\:{to}\:{find}\:{the}\:{value}\:{of}\:\int_{\mathrm{0}} ^{\infty} \:\:\frac{\left(\mathrm{1}−{e}^{−{x}} \right){sinx}}{{x}^{\mathrm{2}} }{dx}\:. \\ $$ Commented by abdo…
Question Number 99504 by pticantor last updated on 21/Jun/20 Answered by Ar Brandon last updated on 21/Jun/20 $$\mathrm{Posons}\:\mathrm{x}=\sqrt{\mathrm{2}}\mathrm{tan}\theta\Rightarrow\mathrm{dx}=\sqrt{\mathrm{2}}\mathrm{sec}^{\mathrm{2}} \theta\mathrm{d}\theta \\ $$$$\mathcal{I}=\int\frac{\sqrt{\mathrm{2}}\mathrm{sec}^{\mathrm{2}} \theta}{\left(\mathrm{2tan}^{\mathrm{2}} \theta+\mathrm{2}\right)^{\mathrm{2}} }\mathrm{d}\theta=\frac{\mathrm{1}}{\mathrm{2}\sqrt{\mathrm{2}}}\int\frac{\mathrm{1}}{\mathrm{sec}^{\mathrm{2}} \theta}\mathrm{d}\theta=\frac{\mathrm{1}}{\mathrm{2}\sqrt{\mathrm{2}}}\int\mathrm{cos}^{\mathrm{2}}…
Question Number 99496 by ~blr237~ last updated on 21/Jun/20 $${convergence}\:{radius}\:{of}\:\:\underset{{n}\in\mathbb{N}} {\sum}\:\mathrm{2}^{{n}} {z}^{{n}!} \: \\ $$ Answered by mathmax by abdo last updated on 21/Jun/20 $$\mathrm{u}_{\mathrm{n}}…