Hard improper integrals Define $\alpha$, such that the following integrals converges: 
$$I_1 = \int_{0}^{\infty} (x + \frac{1}{x})^{\alpha} \ln (1+x^{- 3 \alpha}) d x, \\
I_2 = \int_0^1 x^{\alpha}(1-x)^{\beta} \ln x d x , \\
I_3 = \int_0^{\infty} \frac{|\sin x|}{e^{x^2 \sin^2 (x)}} d x, \\
I_4 = \int_0^{\infty}\frac{d x}{1 + x^{\alpha} \sin x^2}.  $$
May you suggest some ideas, please? Can't solve them for about a week...
 A: The first two:
$I_1$: from $\ln(1+x)\leq x$, $$\left(x+\frac1x\right)^\alpha\ln(1+x^{-3\alpha})\leq \left(x+\frac1x\right)^\alpha \,\frac1{x^{3\alpha}}=\left(1+\frac1{x^2}\right)^\alpha\,\frac{1}{x^{2\alpha}},
$$
so convergence of the integral at $\infty$ is guaranteed if $2\alpha>1$, i.e. $\alpha>1/2$. For $\alpha\leq1/2$, since we only care about big values of $x$ we have 
$$\left(x+\frac1x\right)^\alpha\ln(1+x^{-3\alpha})\geq x^\alpha\,\frac1{2x^{3\alpha}}=\frac1{2x^{2\alpha}}
$$
and the integral will diverge.
At zero, $$\left(x+\frac1x\right)^\alpha\ln(1+x^{-3\alpha})\geq -3\alpha\,\frac {\ln x}{x^\alpha}
$$ and the integral will diverge if $\alpha\geq1$. For $\alpha <1$, $$\left(x+\frac1x\right)^\alpha\ln(1+x^{-3\alpha})\leq 2^\alpha\,\frac {\ln2-3\alpha\ln x}{x^\alpha}$$

$I_2$: we have, for any $\alpha\geq0$,
$$
\left|\int_0^1 x^\alpha(1-x)^{\beta}\,\ln x\,dx\right|\leq\int_0^1|\ln x|\,dx=1,
$$
so the integral converges. For $-1<\alpha<0$, we have 
$$
x^{\alpha}\ln x=x^{\alpha-(\alpha+1)/2}(x^{(\alpha+1)/2}\ln x),
$$
the expression in brackets will be bounded, and the integral converges because $-1<(\alpha-1)/2<0$. For $\alpha\leq-1$ the integral will diverge. The constant $\beta$ deals with the convergence at $1$. When $\beta\geq0$, the integral converges as given by $\alpha$ above. For $\beta<0$, we can see the behaviour by changing variables $x\longmapsto 1-x$, so 
$$
I_2=\int_0^1 (1-x)^\alpha x^\beta\,\ln(1-x)\,dx.
$$
Close to $0$, $$(1-x)^\alpha x^\beta\,\ln(1-x)\simeq x^\beta(-x)=-x^{\beta+1}.$$ So the integral will converge when $\beta+1>-1$, that is $\beta>-2$. It will diverge for $\beta\leq-2$. 
