I want to check if the following improper integrals converge or not. I each integral I wrote my approach.
- $\displaystyle{\int_1^{+\infty}\frac{\ln t+\sin^2t}{t^3}\, dt}$ :
We have that $$\int_1^{+\infty}\frac{\ln t+\sin^2t}{t^3}\, dt=\int_1^{+\infty}\frac{\ln t}{t^3}\, dt+\int_1^{+\infty}\frac{\sin^2t}{t^3}\, dt$$
For the first integral: \begin{align*}\int_1^{+\infty}\frac{\ln t}{t^3}\, dt&=\int_1^{+\infty}\ln t\cdot t^{-3}\, dt=\int_1^{+\infty}\ln t\cdot \left (\frac{t^{-2}}{-2}\right )'\, dt\\ & =\left [\frac{\ln t\cdot t^{-2}}{2}\right ]_1^{+\infty}-\int_1^{+\infty}\frac{1}{t}\cdot \frac{t^{-2}}{-2}\, dt=\left [\frac{\ln t\cdot t^{-2}}{2}\right ]_1^{+\infty}+\int_1^{+\infty}\frac{1}{2t^3}\, dt\\ & =\left [\frac{\ln t\cdot t^{-2}}{2}\right ]_1^{+\infty}-\left [\frac{1}{4t^2}\right ]_1^{+\infty}=\frac{1}{2}\lim_{t\rightarrow +\infty}\frac{\ln t}{t^2}+\frac{1}{4}\\ & =\frac{1}{2}\lim_{t\rightarrow +\infty}\frac{\frac{1}{t}}{2t}+\frac{1}{4}=\frac{1}{2}\lim_{t\rightarrow +\infty}\frac{1}{2t^2}+\frac{1}{4}=\frac{1}{4}\end{align*}
For the second integral: $$\left |\frac{\sin^2t}{t^3}\right |\leq \frac{1}{t^3} \ \text{ and } \ \int_1^{+\infty}\frac{1}{t^3}\, dt=\left [-\frac{1}{2t^2}\right ]_1^{+\infty}=\frac{1}{2}$$ From the comparison test the integral $\int_1^{+\infty}\frac{\sin^2t}{t^3}\, dt$ and so does also the original one. Is everything correct?
- $\displaystyle{\int_0^{+\infty}\frac{t^2\ln t}{(1+t^2)^2}\, dt}$ :
$$\left |\frac{t^2\ln t}{(1+t^2)^2}\right |\leq \frac{t^2\cdot t^{0.5}}{t^4}=\frac{t^{2.5}}{t^4}=t^{-1.5}$$ But this integral doesn't converge at $t=0$, does it?
- $\displaystyle{\int_1^{+\infty}\frac{\left (\ln t\right )^a}{t^{\frac{3}{2}}}, \ a\in \mathbb{R}}$ :
Do we have to take here cases for $a$, if it is positive or negative?
Best Answer
The first one is correct, but its much easier to just use $\ln t +\sin^2 t < t-1 +1 = t$.
For the second, you should distinguish the region near 0. For $t>1$, your estimate is correct and enough. For $0<t<1$, you can use $|\log t| < 1/t$.
For the third, you can show that $\ln t \le C_a t^{\frac1{3a}}$ for every $a>0$ (it follows directly from taking $\ln$s of $r<e^r$ and setting $r=t^{\frac1{3a}}$). If $a= 0$ instead, it should be obvious. For $a<0$, the behavior at infinity is very nice, but since $\log 1=0$, you have a singularity at $t=1$. Since $\ln t = t -1 + O((t-1)^2)$, then $(\log t)^a$ has a singularity like $\frac1{(t-1)^{|a|}}$ at $t=1$. You will therefore need $a>-1$ to have convergence. The integral will diverge for $a\le -1$.