Question Number 75795 by ~blr237~ last updated on 17/Dec/19
$$\mathrm{Find}\:\mathrm{out}\:\:\int_{\mathrm{0}} ^{\infty} \:\frac{\mathrm{argsh}\left(\mathrm{x}\right)}{\mathrm{x}}\mathrm{dx} \\ $$
Commented by mathmax by abdo last updated on 18/Dec/19
![let A =∫_0 ^∞ ((argsh(x))/x)dx ⇒ A =∫_0 ^∞ ((ln(x+(√(1+x^2 ))))/x)dx changement argsh(x)=t give x =sh(t)=((e^t −e^(−t) )/2) ⇒ A =∫_0 ^∞ (t/(sh(t)))ch(t)dt =∫_0 ^∞ ((tch(t))/(sh(t)))dt =∫_0 ^∞ t ((e^t +e^(−t) )/(e^t −e^(−t) )) dt =∫_0 ^∞ t ((1+e^(−2t) )/(1−e^(−2t) ))dt =∫_0 ^∞ (t+te^(−2t) )(Σ_(n=0) ^∞ e^(−2nt) )dt =Σ_(n=0) ^∞ ∫_0 ^∞ t e^(−2nt) dt +Σ_(n=0) ^∞ ∫_0 ^∞ t e^(−(2+2n)t) dt by parts ∫_0 ^∞ t e^(−2nt) dt=[−(t/(2n)) e^(−2nt) ]_0 ^(+∞) −∫_0 ^∞ (−(1/(2n)))e^(−2nt) dt =(1/(2n)) ∫_0 ^∞ e^(−2nt) dt =−(1/(4n^2 )) [e^(−2n) ]_0 ^(+∞) =(1/(4n^2 )) ∫_0 ^∞ t e^(−(2+2n)t) dt =_(t=(u/2)) ∫_0 ^∞ (u/2) e^(−(n+1)t) (du/2) =(1/4)∫_0 ^∞ u e^(−(n+1)u) du =_((n+1)u=z) (1/4)∫_0 ^∞ (z/(n+1))e^(−z) (dz/(n+1)) =(1/(4(n+1)^2 ))∫_0 ^∞ z e^(−z) dz =(1/(4(n+1)^2 )){ [−z e^(−z) ]_0 ^∞ −∫_0 ^∞ (−e^(−z) )} =(1/(4(n+1)^2 )){ [−e^(−z) ]_0 ^(+∞) } =(1/(4(n+1)^2 )) ⇒ A =Σ_(n=0) ^∞ (1/(4n^2 )) +Σ_(n=0) ^∞ (1/(4(n+1)^2 )) so A is divergent ...! or something went wrong..](https://www.tinkutara.com/question/Q75820.png)
$${let}\:{A}\:=\int_{\mathrm{0}} ^{\infty} \:\frac{{argsh}\left({x}\right)}{{x}}{dx}\:\Rightarrow\:{A}\:=\int_{\mathrm{0}} ^{\infty} \frac{{ln}\left({x}+\sqrt{\mathrm{1}+{x}^{\mathrm{2}} }\right)}{{x}}{dx} \\ $$$${changement}\:{argsh}\left({x}\right)={t}\:{give}\:{x}\:={sh}\left({t}\right)=\frac{{e}^{{t}} −{e}^{−{t}} }{\mathrm{2}}\:\Rightarrow \\ $$$${A}\:=\int_{\mathrm{0}} ^{\infty} \:\:\frac{{t}}{{sh}\left({t}\right)}{ch}\left({t}\right){dt}\:\:=\int_{\mathrm{0}} ^{\infty} \:\:\frac{{tch}\left({t}\right)}{{sh}\left({t}\right)}{dt} \\ $$$$=\int_{\mathrm{0}} ^{\infty} \:\:{t}\:\frac{{e}^{{t}} +{e}^{−{t}} }{{e}^{{t}} −{e}^{−{t}} }\:{dt}\:=\int_{\mathrm{0}} ^{\infty} \:{t}\:\frac{\mathrm{1}+{e}^{−\mathrm{2}{t}} }{\mathrm{1}−{e}^{−\mathrm{2}{t}} }{dt} \\ $$$$=\int_{\mathrm{0}} ^{\infty} \:\left({t}+{te}^{−\mathrm{2}{t}} \right)\left(\sum_{{n}=\mathrm{0}} ^{\infty} \:{e}^{−\mathrm{2}{nt}} \right){dt} \\ $$$$=\sum_{{n}=\mathrm{0}} ^{\infty} \int_{\mathrm{0}} ^{\infty} \:{t}\:{e}^{−\mathrm{2}{nt}} \:{dt}\:+\sum_{{n}=\mathrm{0}} ^{\infty} \int_{\mathrm{0}} ^{\infty} \:{t}\:{e}^{−\left(\mathrm{2}+\mathrm{2}{n}\right){t}} \:{dt}\: \\ $$$${by}\:{parts}\:\:\int_{\mathrm{0}} ^{\infty} \:{t}\:{e}^{−\mathrm{2}{nt}} {dt}=\left[−\frac{{t}}{\mathrm{2}{n}}\:{e}^{−\mathrm{2}{nt}} \right]_{\mathrm{0}} ^{+\infty} \:−\int_{\mathrm{0}} ^{\infty} \:\left(−\frac{\mathrm{1}}{\mathrm{2}{n}}\right){e}^{−\mathrm{2}{nt}} \:{dt} \\ $$$$=\frac{\mathrm{1}}{\mathrm{2}{n}}\:\int_{\mathrm{0}} ^{\infty} \:\:{e}^{−\mathrm{2}{nt}} \:{dt}\:=−\frac{\mathrm{1}}{\mathrm{4}{n}^{\mathrm{2}} }\:\left[{e}^{−\mathrm{2}{n}} \right]_{\mathrm{0}} ^{+\infty} =\frac{\mathrm{1}}{\mathrm{4}{n}^{\mathrm{2}} } \\ $$$$\int_{\mathrm{0}} ^{\infty} \:\:{t}\:{e}^{−\left(\mathrm{2}+\mathrm{2}{n}\right){t}} \:{dt}\:=_{{t}=\frac{{u}}{\mathrm{2}}} \:\int_{\mathrm{0}} ^{\infty} \:\:\frac{{u}}{\mathrm{2}}\:{e}^{−\left({n}+\mathrm{1}\right){t}} \:\frac{{du}}{\mathrm{2}} \\ $$$$=\frac{\mathrm{1}}{\mathrm{4}}\int_{\mathrm{0}} ^{\infty} \:\:{u}\:{e}^{−\left({n}+\mathrm{1}\right){u}} \:{du}\:\:=_{\left({n}+\mathrm{1}\right){u}={z}} \:\:\frac{\mathrm{1}}{\mathrm{4}}\int_{\mathrm{0}} ^{\infty} \:\frac{{z}}{{n}+\mathrm{1}}{e}^{−{z}} \:\frac{{dz}}{{n}+\mathrm{1}} \\ $$$$=\frac{\mathrm{1}}{\mathrm{4}\left({n}+\mathrm{1}\right)^{\mathrm{2}} }\int_{\mathrm{0}} ^{\infty} \:{z}\:{e}^{−{z}} \:{dz}\:=\frac{\mathrm{1}}{\mathrm{4}\left({n}+\mathrm{1}\right)^{\mathrm{2}} }\left\{\:\left[−{z}\:{e}^{−{z}} \right]_{\mathrm{0}} ^{\infty} −\int_{\mathrm{0}} ^{\infty} \:\left(−{e}^{−{z}} \right)\right\} \\ $$$$=\frac{\mathrm{1}}{\mathrm{4}\left({n}+\mathrm{1}\right)^{\mathrm{2}} }\left\{\:\left[−{e}^{−{z}} \right]_{\mathrm{0}} ^{+\infty} \right\}\:=\frac{\mathrm{1}}{\mathrm{4}\left({n}+\mathrm{1}\right)^{\mathrm{2}} }\:\Rightarrow \\ $$$${A}\:=\sum_{{n}=\mathrm{0}} ^{\infty} \:\:\frac{\mathrm{1}}{\mathrm{4}{n}^{\mathrm{2}} }\:+\sum_{{n}=\mathrm{0}} ^{\infty} \:\frac{\mathrm{1}}{\mathrm{4}\left({n}+\mathrm{1}\right)^{\mathrm{2}} }\:\:\:{so}\:{A}\:{is}\:{divergent}\:…! \\ $$$${or}\:{something}\:{went}\:{wrong}.. \\ $$