Matematické Fórum

stuart clark · 02. 03. 2019 10:34

Finding $\int \frac{1}{x}\prod^{\infty}_{i=1}\bigg(1-\tan^2 \frac{x}{2^i}\bigg)dx\cdot \frac{1}{\ln\bigg(\sum^{\infty}_{\alpha =1}\bigg[\int^{\infty}_{0}\frac{x^{\alpha-1}\cdot e^{-\frac{1}{n}}}{n\cdot ((\alpha-1)!)^2}dx\bigg]\bigg)}$

laszky · 25. 05. 2019 02:44

↑ stuart clark:

Hi.

Is the first integral definite or indefinite?
Is $n$ a parameter, or is it just a missprint?

For the first integral I found, that since there holds
$1-\tan^2\frac{x}{2^i} = \frac{\cos^2\frac{x}{2^i}-\sin^2\frac{x}{2^i}}{\cos^2\frac{x}{2^i}} = \frac{\cos\frac{x}{2^{i-1}}}{\cos^2\frac{x}{2^i}}$ ,

we have
$\prod_{i=1}^{\infty}\left(1-\tan^2\frac{x}{2^i}\right) = \prod_{i=1}^{\infty}\frac{\cos\frac{x}{2^{i-1}}}{\cos^2\frac{x}{2^i}} = \frac{\cos x}{\prod_{i=1}^{\infty}\cos\frac{x}{2^i}} = \frac{x\cos x}{\sin x}$

and hence
$\int\frac{1}{x}\prod_{i=1}^{\infty}\left(1-\tan^2\frac{x}{2^i}\right)\,\mathrm{d}x = \int \frac{\cos x}{\sin x}\,\mathrm{d}x = \ln|\sin x| + C$

stuart clark · 25. 05. 2019 11:13

Thanks Laszky actually it is given by someone to me and i also seems that it is wrong.

Matematické Fórum

#1 02. 03. 2019 10:34

nested summation with integration

#2 25. 05. 2019 02:44

Re: nested summation with integration

#3 25. 05. 2019 11:13

Re: nested summation with integration

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