help-please-1-1-t-sin-1-1-t-dt-I-can-show-that-it-is-equal-to-n-0-1-n-2n-1-4-n-n-2-2n-1-but-I-cant-calculate-it-

Question Number 170349 by antoineop last updated on 21/May/22

$$$$$$helpplease$$$${\stackrel{\mathrm{\infty }}{\int }}_1(\frac{1}{t}-{\sin }^{-1}\left(\frac{1}{t}\right))dt$$$$Icanshowthatitisequalto$$$$-\sum _{n⩾0}{(-1)}^n\frac{(2n-1)!}{4^n{(n!)}^2(2n+1)}$$$$butIcantcalculateit\dots$$

Answered by aleks041103 last updated on 21/May/22

$$\frac{1}{\sqrt{1-x^2}}={(1+(-x^2))}^{-1/2}$$$${(1+z)}^a=1+az+\frac{a(a-1)}{2}{z}^2+\frac{a(a-1)(a-2)}{3!}{z}^3+\dots =$$$$=1+\underset{i=1}{\sum ^{\mathrm{\infty }}}\frac{z^i}{i!}\underset{j=0}{\prod ^{i-1}}(a-j)$$$$\frac{1}{\sqrt{1-x^2}}=1+\underset{i=1}{\sum ^{\mathrm{\infty }}}\frac{{(-x^2)}^i}{i!}\underset{j=0}{\prod ^{i-1}}(-\frac{1}{2}-j)=$$$$=1+\underset{i=1}{\sum ^{\mathrm{\infty }}}\frac{{(-1)}^i{x}^{2i}}{i!}\underset{j=0}{\prod ^{i-1}}(-\frac{1}{2})(2j+1)=$$$$\underset{j=0}{\prod ^{i-1}}(-\frac{1}{2})(2j+1)={(-\frac{1}{2})}^i(1.3.\dots .(2i-1))$$$$1.3.\dots .(2i-1)=\frac{\left(2i\right)!}{2^{i}i!}$$$$\frac{1}{\sqrt{1-x^2}}=1+\underset{i=1}{\sum ^{\mathrm{\infty }}}\frac{\left(2i\right)!}{4^i{(i!)}^2}{x}^{2i}=\underset{i=0}{\sum ^{\mathrm{\infty }}}\frac{\left(2i\right)!}{4^i{(i!)}^2}{x}^{2i}$$$${sin}^{-1}\left(x\right)={\int }_0^x\frac{dx}{\sqrt{1-x^2}}=$$$$=\underset{i=0}{\sum ^{\mathrm{\infty }}}\frac{\left(2i\right)!}{4^i{(i!)}^2}{\int }_0^x{x}^{2i}dx=\underset{i=0}{\sum ^{\mathrm{\infty }}}\frac{\left(2i\right)!}{4^i(2i+1){(i!)}^2}{x}^{2i+1}$$$$\frac{1}{t}-{sin}^{-1}\left(t\right)=$$$$=\frac{1}{t}-\underset{i=0}{\sum ^{\mathrm{\infty }}}\frac{\left(2i\right)!}{4^i(2i+1){(i!)}^2}\frac{1}{t^{2i+1}}=$$$$=-\underset{i=1}{\sum ^{\mathrm{\infty }}}\frac{\left(2i\right)!}{4^i(2i+1){(i!)}^2}\frac{1}{t^{2i+1}}$$$$\Rightarrow {\int }_1^{\mathrm{\infty }}(\frac{1}{t}-asin\left(\frac{1}{t}\right))dt=$$$$=-\underset{i=1}{\sum ^{\mathrm{\infty }}}\frac{\left(2i\right)!}{4^i(2i+1){(i!)}^2}{\int }_1^{\mathrm{\infty }}{t}^{-2i-1}dt=$$$$=-\underset{i=1}{\sum ^{\mathrm{\infty }}}\frac{\left(2i\right)!}{4^i(2i+1){(i!)}^2}{\left(\frac{t^{-2i}}{-2i}\right]}_1^{\mathrm{\infty }}=$$$$=-\underset{i=1}{\sum ^{\mathrm{\infty }}}\frac{(2i-1)!}{4^i(2i+1){(i!)}^2}$$

Commented by aleks041103 last updated on 21/May/22

Commented by antoineop last updated on 21/May/22

Yep thats what I did, thanks ! (I'm sorry my English is really bad) but I'm supposed to calculate this integral, so, I'm supposed to calculate this sum and I don't know how..

Commented by aleks041103 last updated on 21/May/22

$$Themainpartofthesolutionis$$$$findingaseriesrepresentationfor$$$$theinversesinefunction.$$$$Afterthat{it}^{\prime }seasy.$$

Commented by antoineop last updated on 21/May/22

My exercise just tell me "Justify the existence of this integral and then, calculate it" I dont know how to calculate this sum, I've also tried IBP but .. still impossible

Commented by aleks041103 last updated on 21/May/22

$$Idontthinkthatthereisaclosedform.$$

Commented by antoineop last updated on 21/May/22

That's what I think too, I'll ask my teacher next week and share it if he has a way (of not) to calculate that ! thanks :)

Commented by aleks041103 last updated on 21/May/22

$$Yourtaskwastoprovethattheimtegral$$$$convergesandthencalculateit,which$$$${you}^{\prime }vealreadydonebyfindingthissum$$

Commented by aleks041103 last updated on 21/May/22

$$f\left(x\right)={sin}^{-1}\left(x\right)-x$$$$weanalyzex\in [0,1]$$$$weknowthat$$$$x⩾sin\left(x\right),x⩾0$$$$\Rightarrow {sin}^{-1}x⩾x⩾0$$$$\Rightarrow {sin}^{-1}x-x⩾0$$$$\Rightarrow {\int }_1^{\mathrm{\infty }}-f\left(\frac{1}{t}\right)dt⩽0$$

Commented by antoineop last updated on 22/May/22

yeah but in France in 99% of analysis exercises we are able to calculate directly the sum

Commented by Tawa11 last updated on 08/Oct/22

$$\mathrm{Great}\mathrm{sir}$$

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