What is the integral of cos t/t where limits are 1 to x ?

My book says that it is sinx/x - sin1 + integral (from 1 to x) sin t/ t^{2 }dt for all x>=1.

I can't figure out how .

Please help.

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- Mar 15th 2013, 04:45 AMmrmaaza123Conditional convergence of improper integrals
What is the integral of cos t/t where limits are 1 to x ?

My book says that it is sinx/x - sin1 + integral (from 1 to x) sin t/ t^{2 }dt for all x>=1.

I can't figure out how .

Please help. - Mar 15th 2013, 05:00 AMTheEmptySetRe: Conditional convergence of improper integrals

Use integration by parts.

$\displaystyle u=\frac{1}{t} \implies du=-\frac{1}{t^2}dt$

and

$\displaystyle dv=\cos(t) \implies v=\sin(t)$

This gives

$\displaystyle \int_{1}^{x} \frac{\cos(t)}{t}dt=\frac{\sin(t)}{t}\bigg|_{t=1}^ {t=x}+\int_{1}^x\frac{\sin(t)}{t^2}dt$ - Mar 15th 2013, 05:28 AMmrmaaza123Re: Conditional convergence of improper integrals
Oh, yes. Thank you, i don't really know how i overlooked that !