How can I prove this? I know why it's the case, but I can't prove it.
Give all $\alpha \in \mathbb R$ such that:
a) $\displaystyle\int_a^b x^\alpha dx$ is convergent
b) $\displaystyle\int_1^\infty x^\alpha dx$ is convergent
How can I prove this? I know why it's the case, but I can't prove it.
Give all $\alpha \in \mathbb R$ such that:
a) $\displaystyle\int_a^b x^\alpha dx$ is convergent
b) $\displaystyle\int_1^\infty x^\alpha dx$ is convergent
I will do part b) in detail:
Here, you have an improper integral, since the interval of integration has infinite length. By definition, the integral $\int_1^\infty x^\alpha\, dx$ is convergent if and only if $ \tag{0}\lim_{b\rightarrow \infty}\int_1^b x^\alpha\, dx $ exists (in the finite sense); in which case we say the integral converges to the value of the limit. Note that you evaluate the integral above first; this will give you an expression in $b$. Then you take the limit or demonstrate that it does not exist.
The picture below, which shows the scenario for $\alpha<-1$ and in which the gray shaded region represents $\int_1^b x^\alpha\,dx$, may help one to see why we define $\int_1^\infty x^\alpha\,dx$ in this way:
Now on to computing the required limit. We need to determine when the limit in $(0)$ exists.
Of course, this will likely depend on the particular value of $\alpha$.
Let's deal with a special case first: $\alpha=-1$.
In this case, we have $ \lim_{b\rightarrow\infty}\int_1^b x^{-1}\, dx =\lim_{b\rightarrow\infty} \ln |x|\bigl|_1^b =\lim_{b\rightarrow\infty} \ln b=\infty. $ Thus, for $\alpha=-1$, the improper integral $\int_1^\infty x^\alpha\,dx$ does not converge.
If $\alpha\ne -1$, then: $\tag{1}\eqalign{ \lim_{b\rightarrow\infty} \int_1^b x^{\alpha}\, dx =\lim_{b\rightarrow\infty}{ {x^{\alpha+1}\over \alpha+1} } \Bigl|_1^b =\lim_{b\rightarrow\infty}\Bigl[{ {b^{\alpha+1}\over \alpha+1} - {1\over \alpha+1} }\Bigr]. } $ To evaluate the limit in the right hand side of $(1)$, we consider two cases:
Case 1) $\alpha>-1$.
In this case, the exponent $\alpha+1>0$; whence $\lim\limits_{b\rightarrow\infty} b^{1+\alpha}=\infty$. Thus, evaluating the limit appearing in the right hand side of $(1)$: $ \lim_{b\rightarrow\infty}\Bigl[{ {b^{\alpha+1}\over \alpha+1} - {1\over \alpha+1} }\Bigr]=\infty;$ and so, the improper integral $\int_1^\infty x^\alpha\,dx$ does not converge for $\alpha>-1$.
Case 2) $\alpha<-1$.
In this case, the exponent $\alpha+1<0$; whence $\lim\limits_{b\rightarrow\infty} b^{1+\alpha}=0$. Thus, evaluating the limit appearing in the right hand side of $(1)$: $ \lim_{b\rightarrow\infty}\Bigl[{ {b^{\alpha+1}\over \alpha+1} - {1\over \alpha+1} }\Bigr]= {-1\over \alpha+1};$ and so, the improper integral $\int_1^\infty x^\alpha\,dx$ converges to $-1\over 1+\alpha $ for $\alpha<-1$.
Summarizing what we have done: the improper integral $ \tag{2} \int_1^\infty x^\alpha\,dx $ converges if and only if $\alpha<-1$.
Note that the improper integral in $(2)$ is just a "$p$-integral" which is conventionally written $ \tag{3}\int_1^\infty {1\over x^p}\,dx $ The improper integral $(3)$ converges if and only if $p>1$.
For part a), assuming that you meant to write $\int_0^1 x^\alpha\,dx$, you need to recognize that this is also an improper integral when $\alpha<0$. In this case, the $y$-axis is a vertical asymptote of the graph of $y=x^\alpha$, and to determine if the integral converges, you would compute: $ \lim_{a\rightarrow0^+} \int_a^1 x^\alpha\,dx. $