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Question Number 48261 by Abdo msup. last updated on 21/Nov/18
let f(x) =∫_(1/2) ^1 (dt/(2+ch(xt))) 1) find a explicit form of f(x) 2) calculate g(x)=∫_(1/2) ^1 ((tsh(xt))/((2+ch(xt))^2 ))dt 3) find the value of ∫_(1/2) ^1 (dt/(2+ch(3t))) and ∫_(1/2) ^1 ((tsh(2t))/((2+ch(2t))^2 ))dt 4) let u_n =∫_(1/2) ^1 (dt/(2+ch(nt))) study the convergence of Σu_n and Σ(u_n /n) .
$${let}\:{f}\left({x}\right)\:=\int_{\frac{\mathrm{1}}{\mathrm{2}}} ^{\mathrm{1}} \:\:\frac{{dt}}{\mathrm{2}+{ch}\left({xt}\right)} \\ $$$$\left.\mathrm{1}\right)\:{find}\:{a}\:{explicit}\:{form}\:{of}\:{f}\left({x}\right) \\ $$$$\left.\mathrm{2}\right)\:{calculate}\:{g}\left({x}\right)=\int_{\frac{\mathrm{1}}{\mathrm{2}}} ^{\mathrm{1}} \:\:\frac{{tsh}\left({xt}\right)}{\left(\mathrm{2}+{ch}\left({xt}\right)\right)^{\mathrm{2}} }{dt} \\ $$$$\left.\mathrm{3}\right)\:{find}\:{the}\:{value}\:{of}\:\int_{\frac{\mathrm{1}}{\mathrm{2}}} ^{\mathrm{1}} \:\:\:\frac{{dt}}{\mathrm{2}+{ch}\left(\mathrm{3}{t}\right)}\:{and}\:\int_{\frac{\mathrm{1}}{\mathrm{2}}} ^{\mathrm{1}} \:\:\:\frac{{tsh}\left(\mathrm{2}{t}\right)}{\left(\mathrm{2}+{ch}\left(\mathrm{2}{t}\right)\right)^{\mathrm{2}} }{dt} \\ $$$$\left.\mathrm{4}\right)\:{let}\:{u}_{{n}} \:\:=\int_{\frac{\mathrm{1}}{\mathrm{2}}} ^{\mathrm{1}} \:\:\frac{{dt}}{\mathrm{2}+{ch}\left({nt}\right)}\:\:{study}\:{the}\:{convergence}\:{of}\:\Sigma{u}_{{n}} \\ $$$${and}\:\Sigma\frac{{u}_{{n}} }{{n}}\:. \\ $$
Commented by maxmathsup by imad last updated on 22/Nov/18
1) we have for x ≠0f(x)=∫_(1/2) ^1 (dt/(2+ch(xt))) =_(xt=u) ∫_(x/2) ^x (du/(x(2+ch(u))) =(1/x) ∫_(x/2) ^x (du/(2+((e^u +e^(−u) )/2))) =(2/x) ∫_(x/2) ^x (du/(4 +e^u +e^(−u) )) =_(e^u =α) (2/x) ∫_e^(x/2) ^e^x (dα/(α(4+α +α^(−1) ))) =(2/x) ∫_e^(x/2) ^e^x (dα/(α^2 +4α +1)) =(2/x) ∫_e^(x/2) ^e^x (dα/((α+2)^2 −3)) =(2/x) ∫_e^(x/2) ^e^x (dα/((α−1)(α+5))) =(1/(3x)) ∫_e^(x/2) ^e^x {(1/(α−1)) −(1/(α+5))}dα = (1/(3x))[ln∣((α−1)/(α+5))∣]_e^(x/2) ^e^x =(1/(3x)){ln∣((e^x −1)/(e^x +5))∣−ln∣((e^(x/2) −1)/(e^(x/2) +5))∣}
$$\left.\mathrm{1}\right)\:{we}\:{have}\:{for}\:{x}\:\neq\mathrm{0}{f}\left({x}\right)=\int_{\frac{\mathrm{1}}{\mathrm{2}}} ^{\mathrm{1}} \:\:\frac{{dt}}{\mathrm{2}+{ch}\left({xt}\right)}\:=_{{xt}={u}} \:\:\int_{\frac{{x}}{\mathrm{2}}} ^{{x}} \:\:\:\frac{{du}}{{x}\left(\mathrm{2}+{ch}\left({u}\right)\right.} \\ $$$$=\frac{\mathrm{1}}{{x}}\:\:\int_{\frac{{x}}{\mathrm{2}}} ^{{x}} \:\:\:\frac{{du}}{\mathrm{2}+\frac{{e}^{{u}} \:+{e}^{−{u}} }{\mathrm{2}}}\:=\frac{\mathrm{2}}{{x}}\:\int_{\frac{{x}}{\mathrm{2}}} ^{{x}} \:\:\:\:\frac{{du}}{\mathrm{4}\:+{e}^{{u}} \:+{e}^{−{u}} }\:=_{{e}^{{u}} =\alpha} \:\frac{\mathrm{2}}{{x}}\:\int_{{e}^{\frac{{x}}{\mathrm{2}}} } ^{{e}^{{x}} } \:\:\frac{{d}\alpha}{\alpha\left(\mathrm{4}+\alpha\:+\alpha^{−\mathrm{1}} \right)} \\ $$$$=\frac{\mathrm{2}}{{x}}\:\int_{{e}^{\frac{{x}}{\mathrm{2}}} } ^{{e}^{{x}} } \:\:\:\:\frac{{d}\alpha}{\alpha^{\mathrm{2}} \:+\mathrm{4}\alpha\:+\mathrm{1}}\:=\frac{\mathrm{2}}{{x}}\:\int_{{e}^{\frac{{x}}{\mathrm{2}}} } ^{{e}^{{x}} } \:\:\:\:\frac{{d}\alpha}{\left(\alpha+\mathrm{2}\right)^{\mathrm{2}} \:−\mathrm{3}}\:=\frac{\mathrm{2}}{{x}}\:\int_{{e}^{\frac{{x}}{\mathrm{2}}} } ^{{e}^{{x}} } \:\:\frac{{d}\alpha}{\left(\alpha−\mathrm{1}\right)\left(\alpha+\mathrm{5}\right)} \\ $$$$=\frac{\mathrm{1}}{\mathrm{3}{x}}\:\int_{{e}^{\frac{{x}}{\mathrm{2}}} } ^{{e}^{{x}} } \:\left\{\frac{\mathrm{1}}{\alpha−\mathrm{1}}\:−\frac{\mathrm{1}}{\alpha+\mathrm{5}}\right\}{d}\alpha\:=\:\frac{\mathrm{1}}{\mathrm{3}{x}}\left[{ln}\mid\frac{\alpha−\mathrm{1}}{\alpha+\mathrm{5}}\mid\right]_{{e}^{\frac{{x}}{\mathrm{2}}} } ^{{e}^{{x}} } \\ $$$$=\frac{\mathrm{1}}{\mathrm{3}{x}}\left\{{ln}\mid\frac{{e}^{{x}} −\mathrm{1}}{{e}^{{x}} \:+\mathrm{5}}\mid−{ln}\mid\frac{{e}^{\frac{{x}}{\mathrm{2}}} −\mathrm{1}}{{e}^{\frac{{x}}{\mathrm{2}}} \:+\mathrm{5}}\mid\right\} \\ $$
Commented by maxmathsup by imad last updated on 22/Nov/18
2) we have f^′ (x) =∫_(1/2) ^1 (∂/∂x)((dt/(2+ch(xt)))) = ∫_(1/2) ^1 ((−t sh(xt))/((2+ch(xt))^2 ))dt ⇒ ∫_(1/2) ^1 ((tsh(xt))/((2+ch(xt)^2 )))dt =−f^′ (x) but f^′ (x)=−(1/(3x^2 )){ ln∣((e^x −1)/(e^x +5))∣−ln∣((e^(x/2) −1)/(e^(x/2) +5))∣ +(1/(3x)){(e^x /(e^x −1)) −(e^x /(e^x +5)) −(1/2) (e^(x/2) /e^((x/(2 ))−1) ) +(1/2)(e^(x/2) /(e^(x/2) +5))} ⇒g(x)=−f^′ (x)
$$\left.\mathrm{2}\right)\:{we}\:{have}\:{f}^{'} \left({x}\right)\:=\int_{\frac{\mathrm{1}}{\mathrm{2}}} ^{\mathrm{1}} \:\frac{\partial}{\partial{x}}\left(\frac{{dt}}{\mathrm{2}+{ch}\left({xt}\right)}\right)\:=\:\int_{\frac{\mathrm{1}}{\mathrm{2}}} ^{\mathrm{1}} \frac{−{t}\:{sh}\left({xt}\right)}{\left(\mathrm{2}+{ch}\left({xt}\right)\right)^{\mathrm{2}} }{dt}\:\Rightarrow \\ $$$$\int_{\frac{\mathrm{1}}{\mathrm{2}}} ^{\mathrm{1}} \:\:\frac{{tsh}\left({xt}\right)}{\left(\mathrm{2}+{ch}\left({xt}\right)^{\mathrm{2}} \right)}{dt}\:=−{f}^{'} \left({x}\right)\:{but}\: \\ $$$${f}^{'} \left({x}\right)=−\frac{\mathrm{1}}{\mathrm{3}{x}^{\mathrm{2}} }\left\{\:{ln}\mid\frac{{e}^{{x}} −\mathrm{1}}{{e}^{{x}} \:+\mathrm{5}}\mid−{ln}\mid\frac{{e}^{\frac{{x}}{\mathrm{2}}} −\mathrm{1}}{{e}^{\frac{{x}}{\mathrm{2}}} \:+\mathrm{5}}\mid\right. \\ $$$$+\frac{\mathrm{1}}{\mathrm{3}{x}}\left\{\frac{{e}^{{x}} }{{e}^{{x}} −\mathrm{1}}\:−\frac{{e}^{{x}} }{{e}^{{x}} \:+\mathrm{5}}\:−\frac{\mathrm{1}}{\mathrm{2}}\:\frac{{e}^{\frac{{x}}{\mathrm{2}}} }{{e}^{\frac{{x}}{\mathrm{2}\:}−\mathrm{1}} }\:+\frac{\mathrm{1}}{\mathrm{2}}\frac{{e}^{\frac{{x}}{\mathrm{2}}} }{{e}^{\frac{{x}}{\mathrm{2}}} \:+\mathrm{5}}\right\}\:\Rightarrow{g}\left({x}\right)=−{f}^{'} \left({x}\right)\: \\ $$$$ \\ $$
Commented by maxmathsup by imad last updated on 22/Nov/18
3) ∫_(1/2) ^1 (dt/(2+ch(3t))) =f(3)=(1/9){ln∣((e^3 −1)/(e^3 +5))∣−ln∣((e^(3/2) −1)/(e^(3/2) +5))∣
$$\left.\mathrm{3}\right)\:\int_{\frac{\mathrm{1}}{\mathrm{2}}} ^{\mathrm{1}} \:\:\frac{{dt}}{\mathrm{2}+{ch}\left(\mathrm{3}{t}\right)}\:={f}\left(\mathrm{3}\right)=\frac{\mathrm{1}}{\mathrm{9}}\left\{{ln}\mid\frac{{e}^{\mathrm{3}} −\mathrm{1}}{{e}^{\mathrm{3}} \:+\mathrm{5}}\mid−{ln}\mid\frac{{e}^{\frac{\mathrm{3}}{\mathrm{2}}} −\mathrm{1}}{{e}^{\frac{\mathrm{3}}{\mathrm{2}}} \:+\mathrm{5}}\mid\right. \\ $$$$ \\ $$
Commented by maxmathsup by imad last updated on 22/Nov/18
∫_(1/2) ^1 ((tsh(2t))/((2+ch(2t))^2 ))dt =g(2) =−f^′ (2) =(1/(12)){ln∣((e^2 −1)/(e^2 +5))∣−ln∣((e−1)/(e+5))∣} +(1/6){ (e^2 /(e^2 −1)) −(e^2 /(e^2 +5)) −(1/2) (e/(e−1)) +(1/2) (e/(e+5))}
$$\int_{\frac{\mathrm{1}}{\mathrm{2}}} ^{\mathrm{1}} \:\:\frac{{tsh}\left(\mathrm{2}{t}\right)}{\left(\mathrm{2}+{ch}\left(\mathrm{2}{t}\right)\right)^{\mathrm{2}} }{dt}\:={g}\left(\mathrm{2}\right)\:=−{f}^{'} \left(\mathrm{2}\right)\:=\frac{\mathrm{1}}{\mathrm{12}}\left\{{ln}\mid\frac{{e}^{\mathrm{2}} −\mathrm{1}}{{e}^{\mathrm{2}} \:+\mathrm{5}}\mid−{ln}\mid\frac{{e}−\mathrm{1}}{{e}+\mathrm{5}}\mid\right\} \\ $$$$+\frac{\mathrm{1}}{\mathrm{6}}\left\{\:\frac{{e}^{\mathrm{2}} }{{e}^{\mathrm{2}} −\mathrm{1}}\:−\frac{{e}^{\mathrm{2}} }{{e}^{\mathrm{2}} \:+\mathrm{5}}\:−\frac{\mathrm{1}}{\mathrm{2}}\:\frac{{e}}{{e}−\mathrm{1}}\:+\frac{\mathrm{1}}{\mathrm{2}}\:\frac{{e}}{{e}+\mathrm{5}}\right\} \\ $$

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