What Value of Alpha Ensures the Convergence of This Integral?

[(@apache@)]

Homework Statement

what coefficient $$\alpha$$ this integral is convergent for?:

$$\int_{0}^{\infty}\frac{e^{x \left|\sin x \right|}}{x^{\alpha}} dx$$

Homework Equations

The Attempt at a Solution

I've started with this: I = I₁ + I₂ + I₃, where
$$I_1=\int_{0}^{\epsilon}$$

$$I_2=\int_{\epsilon}^{M}$$

$$I_2=\int_{M}^{\infty}$$

so, I₂ is clear, but I have big problem with the third one, it looks like the divergent, but I'm not really sure, can you help me?
Is it correct that $$I_1 \sim \frac{1}{x^{\alpha}}$$?

 

[Mark44]

You don't need three integrals: two will be enough. For the first one the limits of integration should be epsilon (lower) and b. For the second one, the limits should be b (lower) and M. Both integrals are improper, but for different reasons, and both need to be evaluated using limits. The first integral is evaluated at b and epsilon, and the limit taken as epsilon goes to 0+. The second integral is evaluated at M and b, and the limit taken as M goes to infinity.

 

[(@apache@)]

I think both suggestion to solve this looks similar, because in the same way you must solve the limit in the infinity and I consider it quite difficult. Could you suggest me some more precise way of solution. I was trying to make the series from that. I mean: $$\sum_{n=M}^{\infty}\int_{n\pi}^{(n+1)\pi}$$ ant then I restricted it by high limit and then I was able to integrated when in the end I got some kind of harmonic series. So to sum up $$\alpha \in (0;1)$$ (according to first integral around zero).
Thanks for your time.

 

[Mark44]

I think you need to work on the integrand.
1 <= e^x|sin(x)| <= e^x, so the integrand is bounded by 1/x^$\alpha$ and e^x/x^$\alpha$. Can you find a value for alpha for which the integrand is small enough that it converges?

 

[(@apache@)]

In the first one I´m quite sure that [tex]\alpha [\tex] should be <1.<br /> In the second one I get into the blind alley, neverthless I think we only one option, so we must restrict for some ending M, then we will get the high bound looks like that: [tex]\leq e^{M}\frac{1}{x^{\alpha}}[\tex], finally $$\alpha [\tex] would be >1. But it's very strange resul, isn`t it? Thanks again to everyone$$[/tex][/tex]

 

[Mark44]

In the first one I´m quite sure that $$\alpha [\tex] should be <1.$$

[tex]I don't think so. If [itex]\alpha[\itex] < 1, then x^{$\alpha [\itex]$}[itex]will be small. Since it's in the denominator, the overall expression will be larger than if [itex]\alpha [\itex] were greater than 1.<br /> <blockquote data-attributes="" data-quote="(@apache@)" data-source="post: 2462514" cite="https://www.physicsforums.com/goto/post?id=2462514" class="bbCodeBlock bbCodeBlock--expandable bbCodeBlock--quote js-expandWatch"> <div class="bbCodeBlock-title"> (@apache@) said: </div> <div class="bbCodeBlock-content"> <div class="bbCodeBlock-expandContent js-expandContent "> In the second one I get into the blind alley, neverthless I think we only one option, so we must restrict for some ending M, then we will get the high bound looks like that: [tex]\leq e^{M}\frac{1}{x^{\alpha}}[\tex], finally $$\alpha [\tex] would be >1. But it's very strange resul, isn`t it? Thanks again to everyone$$[/tex] </div> </div> </blockquote>[tex][tex] <br /> My sense is that this integral diverges no matter what [itex]\alpha [\itex] is. The exponential function grows large so fast that division by x^{$\alpha [\itex]$}$won't make any difference.$[/itex][/tex][/tex][/itex][/itex][/itex][/tex]

 

[(@apache@)]

I was optimistic and I believed it could convergent.
Thanks fot your patience
See you