Integral Convergence Homework Solutions

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Mattofix
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Homework Statement



i)
[tex]\int[/tex][tex]^{\infty}_{}[/tex] (cos x)/(x +exp(x)) dx
[tex]_{ 0}[/tex]

ii)
[tex]\int[/tex][tex]^{\infty}_{}[/tex] (x + [tex]\sqrt{x})^{-1}[/tex]dx
[tex]_{ 1}[/tex]

iii)
[tex]\int[/tex][tex]^{\infty}_{}[/tex] (1 + x[tex]^{3}[/tex])[tex]^{-1/2}[/tex] dx
[tex]_{ 1}[/tex]

The Attempt at a Solution

i) x +exp(x) [tex]\geq[/tex] 1

-1[tex]\leq[/tex]cos x [tex]\leq[/tex]1

-1[tex]\leq[/tex](cos x)/(x +exp(x))[tex]\leq[/tex]1/(x +exp(x))

then do i have to compare something to something knowing for certain that something convereges?

For ii) + iii) please can someone nudge me in the right direction.
 
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for the first one use the fact that cosx<1 for x from zero to infinity, also use the fact that 1/(x+exp(x))<exp(-x), then try to show that exp(-x) converges, so there is a theorem i guess, i am not sure how exactly it goes but i think it say that if

f(x)<g(x), then also

integ (from a to x)f(x)dx<integ(from a to x) g(x), then if the right hand sided integral converges say to a nr M, then it means that integ (from a to x)f(x)dx<M, so it means that this function is upper bounded so it also must have a precise upper bound, hence the limit also must exist as x-->infinity, which actually tells us that the integral
[tex]\int_0^{\infty} \frac{cos x}{x+e^{x}}dx[/tex] converges
 
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For the other two i believe you can find the antiderivatives of those functions, and see whether they converge or not!


[tex]\int_0^{\infty}\frac{dx}{x+\sqrt x}=\lim_{b\rightarrow\infty} \int_0^{b}\frac{dx}{x+\sqrt x}[/tex], now let [tex]x=t^{2} => dx=2tdt, <br /> t=\sqrt x[/tex] [tex]x=0 =>t=0, when, x=b => t=\sqrt b[/tex]
[tex]\lim_{b\rightarrow\infty} \int_0^{\sqrt b}\frac{2tdt}{t^{2}+t}=2\lim_{b\rightarrow\infty}\int_0^{\sqrt b}\frac{dt}{1+t}[/tex],
 
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I think also for the iii) you will be able to find an antiderivative in terms of an el. function!