What Does the Improper Integral \int^{\infty}_{o}\frac{sinx}{x} Mean?

twalker40
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1. Let F(x)= \int^{x}_{0} \frac{sint}{t} and f(x) = \frac{sinx}{x}. If x approaches infinity, F(x) approaches \pi/2. So, Explain what does this mean for the improper integral \int^{\infty}_{o}\frac{sinx}{x}



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Explain what does this mean for the improper integral \int^{\infty}_{o}\frac{sinx}{x}


The Attempt at a Solution

 
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What is the definition of
\int_a^\infty f(x) dx[/itex]?
 
So I am guessing that \int^{\infty}_{o}\frac{sinx}{x} converges to pi/2?. The question seems straight forward but my teacher isn't that forgiving, I am thknking there's more to it?
 
Is there any reason why you called the integrand function f? You never used that definiiton in the sequel. Are you sure you copied the question correctly? As it stands it really looks somewhat like senseless:smile:
 
Prove $$\int\limits_0^{\sqrt2/4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx = \frac{\pi^2}{8}.$$ Let $$I = \int\limits_0^{\sqrt 2 / 4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx. \tag{1}$$ The representation integral of ##\arcsin## is $$\arcsin u = \int\limits_{0}^{1} \frac{\mathrm dt}{\sqrt{1-t^2}}, \qquad 0 \leqslant u \leqslant 1.$$ Plugging identity above into ##(1)## with ##u...
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