For which α does the integral ∫ 2 + ∞ e α x ( x − 1 ) α ln ⁡ x d x converge?For which values of α does the integral ∫ 2 + ∞ e α x ( x − 1 ) α ln ⁡ x d xconverge?I'm lost here.For simplicity, let us denote the above integral as I 1 I was able to prove (I noticed that it's kind of standard exercise though) that: I 2 = ∫ 2 + ∞ 1 ( x − 1 ) α ln β ⁡ x I2 converges for α &gt; 1 and ∀ β.converges for α = 1 and β &gt; 1. For β ≤ 1 it diverges.diverges for α &lt; 1 and ∀ β.So my strategy was to use I 2 to prove the converges/divergence of I 1 , by means of inequalities, but I get nothing from this.For instance I did the following:For α &gt; 1 we have that 0 &lt; ∫ 2 + ∞ d x x α ln ⁡ x ≤ ∫ 2 + ∞ e α x ( x − 1 ) α ln ⁡ x d xThat is, for β = 1 0 &lt; I 2 ≤ I 1 From 1. we know that I 2 converges, but this doesn't tell us anything about the convergence of I 1 .The answer given by my textbook is that I 1 converges for α &lt; 0, but I don't know how to arrive at that conclusion.

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For which  α  does the integral       ∫    2          +      ∞                  e              α        x                    (      x      −      1              )                  α                    ln      ⁡      x            d    x converge?For which values of   α does the integral      ∫    2          +      ∞                  e              α        x                    (      x      −      1              )                  α                    ln      ⁡      x            d    xconverge?I&#039;m lost here.For simplicity, let us denote the above integral as       I    1  I was able to prove (I noticed that it&#039;s kind of standard exercise though) that:      I    2    =      ∫    2          +      ∞            1          (      x      −      1              )                  α                            ln                  β                    ⁡      x      I2 converges for   α  &amp;gt;  1 and   ∀     β.converges for   α  =  1 and   β  &amp;gt;  1. For   β  ≤  1 it diverges.diverges for   α  &amp;lt;  1 and   ∀     β.So my strategy was to use       I    2   to prove the converges/divergence of       I    1  , by means of inequalities, but I get nothing from this.For instance I did the following:For   α  &amp;gt;  1 we have that  0  &amp;lt;      ∫    2          +      ∞                          d            x                      x                  α                    ln      ⁡      x        ≤      ∫    2          +      ∞                  e              α        x                    (      x      −      1              )                  α                    ln      ⁡      x            d    xThat is, for   β  =  1  0  &amp;lt;      I    2    ≤      I    1  From 1. we know that       I    2   converges, but this doesn&#039;t tell us anything about the convergence of       I    1  .The answer given by my textbook is that       I    1   converges for   α  &amp;lt;  0, but I don&#039;t know how to arrive at that conclusion.

Cremolinoer · Accepted Answer

The main point is that exponentials are always going to dominate polynomials Specifically, for any   ϵ  &amp;gt;  0 there exists a constant       C    ϵ   such that       e    x    ≥      C    ϵ    (  x  −  1      )    ϵ   for all   x  ≥  2. This implies that for   α  &amp;gt;  0,       e          α      x        ≥      C    ϵ    α    (  1  −  x      )          α      ϵ      , so choosing   ϵ  &amp;lt;  1 such that   0  &amp;lt;  α  (  1  −  ϵ  )  &amp;lt;  1, we find            e              α        x                    (      x      −      1              )        α            log      ⁡      x        ≥      C    ϵ    α        1          (      x      −      1              )                  −          α          (          1          −          ϵ          )                    log      ⁡      x      and so by comparison with       I    2   you know       I    1   diverges. Similarly, if   α  &amp;lt;  0 then       e          α      x              /            2        ≤      C    2          α        (  x  −  1      )          2      α       and hence            e              α        x                    (      x      −      1              )        α            log      ⁡      x        ≤      C    2          α              /            2            1          (      x      −      1              )                  −          α                    log      ⁡      x        ,so by comparison with       I    2   you know       I    1   converges.

hotonglamoz · Accepted Answer

What you&#039;ve shown about the integral w/o the exponential factor is actually somewhat harder conceptually than the integral at hand. If   α  &amp;lt;  0 then the exponential factor is an exponential decay which overpowers any of the other behavior and will make the integral converge. If   α  &amp;gt;  0  ,, it&#039;s an exponential growth that will likewise make it diverge.   α  =  0 you&#039;ve already handled.

"For which alpha does the integral int_2^+inf ( e^(alpha x)/((x−1)^alpha ln x) converge? For which values of α does the integral int_2^+inf ( e^(alpha x)/((x−1)^alpha ln x) dx converge?

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New Questions in Algebra I