Question Number 102894 by bramlex last updated on 11/Jul/20 $$\int\:\sqrt{\mathrm{x}+\sqrt{\mathrm{x}+\sqrt{\mathrm{x}+\sqrt{\mathrm{x}+\sqrt{\mathrm{x}+…}}}}}\:\mathrm{dx}\: \\ $$ Answered by bemath last updated on 11/Jul/20 $${y}=\sqrt{{x}+\sqrt{{x}+\sqrt{{x}+\sqrt{{x}+\sqrt{…}}}}}\: \\ $$$${y}^{\mathrm{2}} \:=\:{x}+{y}\:\Rightarrow{y}^{\mathrm{2}} −{y}−{x}\:=\:\mathrm{0} \\…
Question Number 37357 by math khazana by abdo last updated on 12/Jun/18 $${let}\:{a}>\mathrm{0}\:{b}\:{from}\:{C}\:{and}\:\:{Re}\left({b}\right)>\mathrm{0} \\ $$$$\left.\mathrm{1}\right)\:{calculate}\:\:\int_{−\infty} ^{+\infty} \:\:\:\frac{{b}\:{cos}\left({ax}\right)}{{x}^{\mathrm{2}} \:+{b}^{\mathrm{2}} }{dx} \\ $$$$\left.\mathrm{2}\right)\:{find}\:{the}\:{value}\:{of}\:\:\int_{−\infty} ^{+\infty} \:\:\:\frac{{x}\:{sin}\left({ax}\right)}{{x}^{\mathrm{2}} \:+{b}^{\mathrm{2}} }\:{dx}.…
Question Number 37356 by math khazana by abdo last updated on 12/Jun/18 $${let}\:\:{b}\:\in{C}\:\:{and}\:{Re}\left({b}\right)\:>\mathrm{0}\:{prove}\:{that} \\ $$$$\int_{−\infty} ^{+\infty} \:\:\:\frac{{e}^{{iax}} }{{x}−{ib}}{dx}\:=\mathrm{2}{i}\pi\:{e}^{−{ab}\:\:} \:\:\:{and} \\ $$$$\int_{−\infty} ^{+\infty} \:\:\:\frac{{e}^{{iax}} }{{x}+{ib}}\:{dx}\:=\mathrm{0} \\…
Question Number 37354 by math khazana by abdo last updated on 12/Jun/18 $${let}\:{f}\left({z}\right)\:=\:\:\frac{{z}^{\mathrm{2}} \:+\mathrm{1}}{\left({z}^{\mathrm{2}} \:−\mathrm{1}\right)\left({z}^{\mathrm{2}} \:−\mathrm{4}\right)} \\ $$$${developp}\:{f}\:{at}\:{integr}\:{serie}\:. \\ $$ Commented by math khazana by…
Question Number 37352 by math khazana by abdo last updated on 12/Jun/18 $${let}\:\:{f}\left({z}\right)\:=\:{e}^{−\frac{\mathrm{1}}{{z}^{\mathrm{2}} }} \:\: \\ $$$$\left.\mathrm{1}\right)\:{give}\:{f}\left({z}\right)\:{at}\:{form}\:{of}\:{serie} \\ $$$$\left.\mathrm{2}\right)\:{give}\:\:\int_{\mathrm{1}} ^{\mathrm{2}} {f}\left({z}\right){dz}\:\:\:{at}\:{form}\:{of}\:{serie}\:. \\ $$ Commented by…
Question Number 37353 by math khazana by abdo last updated on 12/Jun/18 $${let}\:{g}\left({z}\right)\:=\frac{{z}}{{e}^{{z}} −\mathrm{1}} \\ $$$${developp}\:{g}\:{at}\:{integr}\:{serie}\:. \\ $$ Commented by prof Abdo imad last updated…
Question Number 37349 by math khazana by abdo last updated on 12/Jun/18 $${calculate}\:\:\:\int_{\mathrm{0}} ^{\mathrm{2}\pi} \:\:\:\:\frac{{dt}}{\mathrm{1}−\mathrm{2}{pcost}\:+{p}^{\mathrm{2}} }\:\:{if}\:\mid{p}\mid<\mathrm{1} \\ $$ Commented by math khazana by abdo last…
Question Number 37350 by math khazana by abdo last updated on 12/Jun/18 $${fond}\:{the}\:{value}\:{of}\:\:\int_{\mathrm{0}} ^{\mathrm{2}\pi} \:\:\:\:\frac{{dt}}{\left({a}\:+{cost}\right)^{\mathrm{2}} }\:\:{with}\:{a}>\mathrm{1}. \\ $$$$ \\ $$ Commented by math khazana by…
Question Number 37347 by math khazana by abdo last updated on 12/Jun/18 $${let}\:{r}\:=\sqrt{{p}^{\mathrm{2}} \:+{q}^{\mathrm{2}} }\:\:\:{p}\:{and}\:{q}\:{from}\:{R}\:\:{and}\:{p}>\mathrm{0}\:\:{q}>\mathrm{0} \\ $$$$\left.\mathrm{1}\right){prove}\:{that}\:\:\int_{\mathrm{0}} ^{+\infty} \:\:{e}^{−{px}} \:\frac{{cos}\left({px}\right)}{\:\sqrt{{x}}}{dx}=\frac{\sqrt{\pi}}{{r}}\sqrt{\frac{{r}+{p}}{\mathrm{2}}} \\ $$$$\left.\mathrm{2}\right)\:\int_{\mathrm{0}} ^{\infty} \:\:{e}^{−{px}} \:\:\frac{{sin}\left({qx}\right)}{\:\sqrt{{x}}}{dx}\:=\frac{\sqrt{\pi}}{{r}}\:\sqrt{\frac{{r}−{p}}{\mathrm{2}}}…
Question Number 37348 by math khazana by abdo last updated on 12/Jun/18 $${calculate}\:\int_{\mathrm{0}} ^{\mathrm{2}\pi} \:\:\:\:\:\frac{{dt}}{{p}\:+{cost}}\:\:{with}\:{p}>\mathrm{1} \\ $$ Commented by prof Abdo imad last updated on…