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Question Number 123898 by mnjuly1970 last updated on 29/Nov/20
... nice calculus... find a series representation for the following integral :: φ=∫_0 ^( 1) cos(h(xln(x))dx
$$\:\:\:\:\:\:\:\:\:…\:{nice}\:\:{calculus}… \\ $$$$\:\:{find}\:\:{a}\:{series}\:{representation} \\ $$$$\:{for}\:\:{the}\:{following}\:{integral}\::: \\ $$$$\:\:\:\:\phi=\int_{\mathrm{0}} ^{\:\mathrm{1}} {cos}\left({h}\left({xln}\left({x}\right)\right){dx}\right. \\ $$$$ \\ $$
Answered by mindispower last updated on 29/Nov/20
∫_0 ^1 ln^a (x)dx=∫_0 ^∞ (−t)^a e^(−t) dt...1 =(−1)^a Γ(a+1).....nice cos(x)=Σ_(n≥0) (((−1)^n x^(2n) )/((2n)!)) ∅=∫_0 ^1 Σ_(n≥0) (((−1)^n )/(2n!)).(hln(x))^(2n) dx =Σ_(n≥0) (((−1)^n )/((2n)!))h^(2n) ∫_0 ^1 ln^(2n) (x)dx....appie 1 a=2n =Σ_(n≥0) (((−1)^n )/((2n)!))h^(2n) .(((−1)^(2n) Γ(2n+1))/) =Σ_(n≥0) (−h^2 )^n =(1/(1+h^2 ))
$$\int_{\mathrm{0}} ^{\mathrm{1}} {ln}^{{a}} \left({x}\right){dx}=\int_{\mathrm{0}} ^{\infty} \left(−{t}\right)^{{a}} {e}^{−{t}} {dt}…\mathrm{1} \\ $$$$=\left(−\mathrm{1}\right)^{{a}} \Gamma\left({a}+\mathrm{1}\right)…..{nice} \\ $$$${cos}\left({x}\right)=\underset{{n}\geqslant\mathrm{0}} {\sum}\frac{\left(−\mathrm{1}\right)^{{n}} {x}^{\mathrm{2}{n}} }{\left(\mathrm{2}{n}\right)!} \\ $$$$\emptyset=\int_{\mathrm{0}} ^{\mathrm{1}} \underset{{n}\geqslant\mathrm{0}} {\sum}\frac{\left(−\mathrm{1}\right)^{{n}} }{\mathrm{2}{n}!}.\left({hln}\left({x}\right)\right)^{\mathrm{2}{n}} {dx} \\ $$$$=\underset{{n}\geqslant\mathrm{0}} {\sum}\frac{\left(−\mathrm{1}\right)^{{n}} }{\left(\mathrm{2}{n}\right)!}{h}^{\mathrm{2}{n}} \:\int_{\mathrm{0}} ^{\mathrm{1}} {ln}^{\mathrm{2}{n}} \left({x}\right){dx}….{appie}\:\mathrm{1}\:{a}=\mathrm{2}{n} \\ $$$$=\underset{{n}\geqslant\mathrm{0}} {\sum}\frac{\left(−\mathrm{1}\right)^{{n}} }{\left(\mathrm{2}{n}\right)!}{h}^{\mathrm{2}{n}} .\frac{\left(−\mathrm{1}\right)^{\mathrm{2}{n}} \Gamma\left(\mathrm{2}{n}+\mathrm{1}\right)}{} \\ $$$$=\underset{{n}\geqslant\mathrm{0}} {\sum}\left(−{h}^{\mathrm{2}} \right)^{{n}} =\frac{\mathrm{1}}{\mathrm{1}+{h}^{\mathrm{2}} } \\ $$
Commented by Mammadli last updated on 29/Nov/20
Commented by Dwaipayan Shikari last updated on 29/Nov/20
lim_(x→1) ((a(a^(x−1) −1))/(x−1))=a.log(a)
$$\underset{{x}\rightarrow\mathrm{1}} {\mathrm{lim}}\frac{{a}\left({a}^{{x}−\mathrm{1}} −\mathrm{1}\right)}{{x}−\mathrm{1}}={a}.{log}\left({a}\right) \\ $$
Commented by mnjuly1970 last updated on 29/Nov/20
thanks for your work sir power.but h:hypbolic
$${thanks}\:{for}\:{your}\:{work} \\ $$$${sir}\:{power}.{but} \\ $$$${h}:{hypbolic} \\ $$
Answered by Dwaipayan Shikari last updated on 29/Nov/20
∫_0 ^1 cos(h(log(x))dx =∫_0 ^1 ((e^(logx) +e^(−logx) )/2)dx =(1/2)∫_0 ^1 Σ_(n=0) ^∞ (((logx)^n )/(n!))+(1/2)∫_0 ^1 Σ_(n=0) ^∞ (((−1)^n (logx)^n )/(n!)) =(1/2)(Σ_(n=0) ^∞ (1/(n!))∫_0 ^1 (logx)^n +Σ_(n=0) ^∞ (((−1)^n )/(n!))∫_0 ^1 (logx)^n ) =(1/2)Σ_(n=0) ^∞ (1/(n!))∫_(−∞) ^0 e^t t^n +(1/2)Σ_(n=0) ^∞ (((−1)^n )/(n!))∫_(−∞) ^0 e^t t^n dt t=−Φ =(1/2)Σ_(n=0) ^∞ (((−1)^n )/(n!))∫_0 ^∞ e^(−Φ) Φ^n dΦ+(1/2)Σ_(n=0) ^∞ (1/(n!))∫_0 ^∞ e^(−Φ) Φ^n dΦ =Σ_(n=0) ^∞ (−1)^n ((Γ(n+1))/(2n!))+Σ_(n=0) ^∞ ((Γ(n+1))/(2n!))
$$\int_{\mathrm{0}} ^{\mathrm{1}} {cos}\left({h}\left({log}\left({x}\right)\right){dx}\right. \\ $$$$=\int_{\mathrm{0}} ^{\mathrm{1}} \frac{{e}^{{logx}} +{e}^{−{logx}} }{\mathrm{2}}{dx} \\ $$$$=\frac{\mathrm{1}}{\mathrm{2}}\int_{\mathrm{0}} ^{\mathrm{1}} \underset{{n}=\mathrm{0}} {\overset{\infty} {\sum}}\frac{\left({logx}\right)^{{n}} }{{n}!}+\frac{\mathrm{1}}{\mathrm{2}}\int_{\mathrm{0}} ^{\mathrm{1}} \underset{{n}=\mathrm{0}} {\overset{\infty} {\sum}}\frac{\left(−\mathrm{1}\right)^{{n}} \left({logx}\right)^{{n}} }{{n}!} \\ $$$$=\frac{\mathrm{1}}{\mathrm{2}}\left(\underset{{n}=\mathrm{0}} {\overset{\infty} {\sum}}\frac{\mathrm{1}}{{n}!}\int_{\mathrm{0}} ^{\mathrm{1}} \left({logx}\right)^{{n}} +\underset{{n}=\mathrm{0}} {\overset{\infty} {\sum}}\frac{\left(−\mathrm{1}\right)^{{n}} }{{n}!}\int_{\mathrm{0}} ^{\mathrm{1}} \left({logx}\right)^{{n}} \right) \\ $$$$=\frac{\mathrm{1}}{\mathrm{2}}\underset{{n}=\mathrm{0}} {\overset{\infty} {\sum}}\frac{\mathrm{1}}{{n}!}\int_{−\infty} ^{\mathrm{0}} {e}^{{t}} {t}^{{n}} +\frac{\mathrm{1}}{\mathrm{2}}\underset{{n}=\mathrm{0}} {\overset{\infty} {\sum}}\frac{\left(−\mathrm{1}\right)^{{n}} }{{n}!}\int_{−\infty} ^{\mathrm{0}} {e}^{{t}} {t}^{{n}} {dt}\:\:\:\:\:\:{t}=−\Phi \\ $$$$=\frac{\mathrm{1}}{\mathrm{2}}\underset{{n}=\mathrm{0}} {\overset{\infty} {\sum}}\frac{\left(−\mathrm{1}\right)^{{n}} }{{n}!}\int_{\mathrm{0}} ^{\infty} {e}^{−\Phi} \Phi^{{n}} {d}\Phi+\frac{\mathrm{1}}{\mathrm{2}}\underset{{n}=\mathrm{0}} {\overset{\infty} {\sum}}\frac{\mathrm{1}}{{n}!}\int_{\mathrm{0}} ^{\infty} {e}^{−\Phi} \Phi^{{n}} {d}\Phi \\ $$$$=\underset{{n}=\mathrm{0}} {\overset{\infty} {\sum}}\left(−\mathrm{1}\right)^{{n}} \frac{\Gamma\left({n}+\mathrm{1}\right)}{\mathrm{2}{n}!}+\underset{{n}=\mathrm{0}} {\overset{\infty} {\sum}}\frac{\Gamma\left({n}+\mathrm{1}\right)}{\mathrm{2}{n}!} \\ $$$$ \\ $$
Commented by mnjuly1970 last updated on 29/Nov/20
thank you so much sir payan . question was corrected : integrand: cos(h(xln(x)))
$${thank}\:{you}\:\:{so}\:{much}\:\:{sir} \\ $$$${payan}\:. \\ $$$${question}\:{was}\:{corrected}\:: \\ $$$${integrand}:\:{cos}\left({h}\left({xln}\left({x}\right)\right)\right) \\ $$$$ \\ $$
Answered by Dwaipayan Shikari last updated on 29/Nov/20
∫_0 ^1 cos(h(xlogx))dx =∫_0 ^1 ((e^(xlogx) +e^(−xlogx) )/2)dx =(1/2)Σ_(n≥0) ^∞ ∫_0 ^1 (((xlogx)^n )/(n!))−(1/2)Σ_(n≥0) (−1)^n ∫_0 ^1 (((xlogx)^n )/(n!)) (t(n+1)=−Φ) =(1/2)Σ_(n≥0) (1/(n!))∫_(−∞) ^0 e^((n+1)t) t^n dt−(1/2)Σ_(n≥0) (−1)^n (1/(n!))∫_(−∞) ^0 e^(t(n+1)) t^n dt =(1/2)Σ_(n≥0) (1/(n!)) (((−1))/((n+1)^(n+1) ))∫_0 ^∞ e^(−Φ) Φ^n dΦ−(1/2)Σ_(n≥0) (((−1)^(2n) )/(n!(n+1)^(n+1) ))∫_0 ^∞ e^(−Φ) Φ^n dΦ =(1/2)Σ_(n≥0) (((−1)^n )/((n+1)^(n+1) ))−(1/2)Σ_(n≥0) (1/((n+1)^((n+1)) )) =(1/2)((1/1^1 )−(1/2^2 )+(1/3^3 )−(1/4^4 )+...−(1/1^1 )−(1/2^2 )−(1/3^3 )−(1/4^4 )+−..) =−((1/2^2 )+(1/4^4 )+(1/6^6 )+(1/8^8 )+....)=−Σ_(n≥1) (1/((2n)^(2n) ))
$$\int_{\mathrm{0}} ^{\mathrm{1}} {cos}\left({h}\left({xlogx}\right)\right){dx} \\ $$$$=\int_{\mathrm{0}} ^{\mathrm{1}} \frac{{e}^{{xlogx}} +{e}^{−{xlogx}} }{\mathrm{2}}{dx} \\ $$$$=\frac{\mathrm{1}}{\mathrm{2}}\underset{{n}\geqslant\mathrm{0}} {\overset{\infty} {\sum}}\int_{\mathrm{0}} ^{\mathrm{1}} \frac{\left({xlogx}\right)^{{n}} }{{n}!}−\frac{\mathrm{1}}{\mathrm{2}}\underset{{n}\geqslant\mathrm{0}} {\sum}\left(−\mathrm{1}\right)^{{n}} \int_{\mathrm{0}} ^{\mathrm{1}} \frac{\left({xlogx}\right)^{{n}} }{{n}!}\:\:\:\:\:\:\left({t}\left({n}+\mathrm{1}\right)=−\Phi\right) \\ $$$$=\frac{\mathrm{1}}{\mathrm{2}}\underset{{n}\geqslant\mathrm{0}} {\sum}\frac{\mathrm{1}}{{n}!}\int_{−\infty} ^{\mathrm{0}} {e}^{\left({n}+\mathrm{1}\right){t}} {t}^{{n}} {dt}−\frac{\mathrm{1}}{\mathrm{2}}\underset{{n}\geqslant\mathrm{0}} {\sum}\left(−\mathrm{1}\right)^{{n}} \frac{\mathrm{1}}{{n}!}\int_{−\infty} ^{\mathrm{0}} {e}^{{t}\left({n}+\mathrm{1}\right)} {t}^{{n}} {dt}\:\:\:\:\:\: \\ $$$$=\frac{\mathrm{1}}{\mathrm{2}}\underset{{n}\geqslant\mathrm{0}} {\sum}\frac{\mathrm{1}}{{n}!}\:\frac{\left(−\mathrm{1}\right)}{\left({n}+\mathrm{1}\right)^{{n}+\mathrm{1}} }\int_{\mathrm{0}} ^{\infty} {e}^{−\Phi} \Phi^{{n}} {d}\Phi−\frac{\mathrm{1}}{\mathrm{2}}\underset{{n}\geqslant\mathrm{0}} {\sum}\frac{\left(−\mathrm{1}\right)^{\mathrm{2}{n}} }{{n}!\left({n}+\mathrm{1}\right)^{{n}+\mathrm{1}} }\int_{\mathrm{0}} ^{\infty} {e}^{−\Phi} \Phi^{{n}} {d}\Phi \\ $$$$=\frac{\mathrm{1}}{\mathrm{2}}\underset{{n}\geqslant\mathrm{0}} {\sum}\frac{\left(−\mathrm{1}\right)^{{n}} }{\left({n}+\mathrm{1}\right)^{{n}+\mathrm{1}} }−\frac{\mathrm{1}}{\mathrm{2}}\underset{{n}\geqslant\mathrm{0}} {\sum}\frac{\mathrm{1}}{\left({n}+\mathrm{1}\right)^{\left({n}+\mathrm{1}\right)} \:\:} \\ $$$$=\frac{\mathrm{1}}{\mathrm{2}}\left(\frac{\mathrm{1}}{\mathrm{1}^{\mathrm{1}} }−\frac{\mathrm{1}}{\mathrm{2}^{\mathrm{2}} }+\frac{\mathrm{1}}{\mathrm{3}^{\mathrm{3}} }−\frac{\mathrm{1}}{\mathrm{4}^{\mathrm{4}} }+…−\frac{\mathrm{1}}{\mathrm{1}^{\mathrm{1}} }−\frac{\mathrm{1}}{\mathrm{2}^{\mathrm{2}} }−\frac{\mathrm{1}}{\mathrm{3}^{\mathrm{3}} }−\frac{\mathrm{1}}{\mathrm{4}^{\mathrm{4}} }+−..\right) \\ $$$$=−\left(\frac{\mathrm{1}}{\mathrm{2}^{\mathrm{2}} }+\frac{\mathrm{1}}{\mathrm{4}^{\mathrm{4}} }+\frac{\mathrm{1}}{\mathrm{6}^{\mathrm{6}} }+\frac{\mathrm{1}}{\mathrm{8}^{\mathrm{8}} }+….\right)=−\underset{{n}\geqslant\mathrm{1}} {\sum}\frac{\mathrm{1}}{\left(\mathrm{2}{n}\right)^{\mathrm{2}{n}} } \\ $$
Commented by mnjuly1970 last updated on 29/Nov/20
peace be upon you sir payan tireless , powerful and unassuming. grateful and god keep you..
$${peace}\:{be}\:{upon}\:{you}\: \\ $$$${sir}\:{payan} \\ $$$${tireless}\:,\:{powerful}\: \\ $$$${and}\:\:{unassuming}.\: \\ $$$${grateful}\:{and}\:{god}\:{keep}\:{you}.. \\ $$
Commented by Dwaipayan Shikari last updated on 29/Nov/20
With pleasure sir! you are really great man
$${With}\:{pleasure}\:{sir}!\:{you}\:{are}\:{really}\:{great}\:{man} \\ $$

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