Question and Answers Forum

All Questions      Topic List

Differentiation Questions

Previous in All Question      Next in All Question      

Previous in Differentiation      Next in Differentiation      

Question Number 146442 by qaz last updated on 13/Jul/21

(d/dn)∣_(n=1) H_n =?

$$\frac{\mathrm{d}}{\mathrm{dn}}\mid_{\mathrm{n}=\mathrm{1}} \mathrm{H}_{\mathrm{n}} =? \\ $$

Answered by mnjuly1970 last updated on 13/Jul/21

= (d/dn)(∫_0 ^( 1) ((1−x^( n) )/(1−x)))=∫_0 ^( 1) ((−x^( n) ln(x ))/(1−x))dx = ∫_0 ^( 1) ((−xln(x))/(1−x)) dx=−∫_0 ^( 1) (((1−x)ln(1−x))/x)dx = li_2 ( 1) +∫_0 ^( 1) ln(1−x)dx = (π^( 2) /6) + ∫_0 ^( 1) ln(x)dx=−1+(π^( 2) /6) ....

$$=\:\frac{{d}}{{dn}}\left(\int_{\mathrm{0}} ^{\:\mathrm{1}} \frac{\mathrm{1}−{x}^{\:{n}} }{\mathrm{1}−{x}}\right)=\int_{\mathrm{0}} ^{\:\mathrm{1}} \frac{−{x}^{\:{n}} {ln}\left({x}\:\right)}{\mathrm{1}−{x}}{dx} \\ $$$$\:\:\:=\:\int_{\mathrm{0}} ^{\:\mathrm{1}} \frac{−{xln}\left({x}\right)}{\mathrm{1}−{x}}\:{dx}=−\int_{\mathrm{0}} ^{\:\mathrm{1}} \frac{\left(\mathrm{1}−{x}\right){ln}\left(\mathrm{1}−{x}\right)}{{x}}{dx} \\ $$$$=\:{li}_{\mathrm{2}} \left(\:\mathrm{1}\right)\:+\int_{\mathrm{0}} ^{\:\mathrm{1}} {ln}\left(\mathrm{1}−{x}\right){dx} \\ $$$$\:=\:\frac{\pi^{\:\mathrm{2}} }{\mathrm{6}}\:+\:\int_{\mathrm{0}} ^{\:\mathrm{1}} {ln}\left({x}\right){dx}=−\mathrm{1}+\frac{\pi^{\:\mathrm{2}} }{\mathrm{6}}\:.... \\ $$

Answered by mnjuly1970 last updated on 13/Jul/21

ψ (n+1 ):= −γ + H_( n) (1/n) + ψ (n) =−γ +H_( n) (d/(d n)) (H n)∣_ = −1 +ψ′(1) = −1 +(π^( 2) /6)= −1+ ζ (2) ...

$$\:\:\psi\:\left({n}+\mathrm{1}\:\right):=\:−\gamma\:+\:\mathrm{H}_{\:{n}} \\ $$$$\:\:\:\:\frac{\mathrm{1}}{{n}}\:+\:\psi\:\left({n}\right)\:=−\gamma\:+\mathrm{H}_{\:{n}} \\ $$$$\:\:\:\:\frac{\mathrm{d}}{\mathrm{d}\:{n}}\:\left(\mathrm{H}\:{n}\right)\underset{} {\mid}=\:−\mathrm{1}\:+\psi'\left(\mathrm{1}\right) \\ $$$$\:\:\:\:\:\:\:\:\:=\:−\mathrm{1}\:+\frac{\pi^{\:\mathrm{2}} }{\mathrm{6}}=\:−\mathrm{1}+\:\zeta\:\left(\mathrm{2}\right)\:... \\ $$

Commented by qaz last updated on 13/Jul/21

thank you sir. I think it is possible to use differential chain rule, same like (∂/∂x)∫_(h(x)) ^(g(x)) f(x,t)dt=∫_(h(x)) ^(g(x)) (∂/∂x)f(x,t)dt+g(x)′f(x,g(x))−h(x)′f(x,h(x)) ,to differential to a seiris summation.....

$$\mathrm{thank}\:\mathrm{you}\:\mathrm{sir}. \\ $$$$\mathrm{I}\:\mathrm{think}\:\mathrm{it}\:\mathrm{is}\:\mathrm{possible}\:\mathrm{to}\:\mathrm{use}\:\mathrm{differential}\:\mathrm{chain}\:\mathrm{rule}, \\ $$$$\mathrm{same}\:\mathrm{like}\:\frac{\partial}{\partial\mathrm{x}}\int_{\mathrm{h}\left(\mathrm{x}\right)} ^{\mathrm{g}\left(\mathrm{x}\right)} \mathrm{f}\left(\mathrm{x},\mathrm{t}\right)\mathrm{dt}=\int_{\mathrm{h}\left(\mathrm{x}\right)} ^{\mathrm{g}\left(\mathrm{x}\right)} \frac{\partial}{\partial\mathrm{x}}\mathrm{f}\left(\mathrm{x},\mathrm{t}\right)\mathrm{dt}+\mathrm{g}\left(\mathrm{x}\right)'\mathrm{f}\left(\mathrm{x},\mathrm{g}\left(\mathrm{x}\right)\right)−\mathrm{h}\left(\mathrm{x}\right)'\mathrm{f}\left(\mathrm{x},\mathrm{h}\left(\mathrm{x}\right)\right) \\ $$$$,\mathrm{to}\:\mathrm{differential}\:\mathrm{to}\:\mathrm{a}\:\mathrm{seiris}\:\mathrm{summation}..... \\ $$

Commented by mnjuly1970 last updated on 13/Jul/21

thank you sir qaz

$${thank}\:{you}\:{sir}\:{qaz} \\ $$

Terms of Service

Privacy Policy

Contact: info@tinkutara.com