# $I(a,b)=\int_0^1 \sin\left(\ln \frac{1}{x}\right)\frac{x^b-x^a}{\ln x}dx,\ b>0,a>0$

$$I(a,b)=\int_0^1 \sin\left(\ln \frac{1}{x}\right)\frac{x^b-x^a}{\ln x}dx,\ b>0,a>0$$

Please help. I can't find a function to see if the function is absolute and uniform convergent using Weierstrass convergence criteria.

Using your definition, we see that there is a natural log in the denominator. An easy way to remove that is simply by using differentiation under the integral. Therefore, we differentiate with respect to $$b$$ because$$\frac {\mathrm dx^b}{\mathrm db}=x^b\log x$$Hence\begin{align*}\mathfrak I'(a,b) & =\int\limits_0^1\mathrm dx\, x^b\sin\left(\log\frac 1x\right)\end{align*}Now make a transformation $$x\mapsto\log\tfrac 1x$$ to get rid of the nested sine - log function. Therefore, the limits change to zero and infinity, so that\begin{align*}\mathfrak{I}'(a,b) & =\int\limits_0^{\infty}\mathrm dx\,e^{x(1+b)}\sin x\\ & =\frac 1{1+(1+b)^2}\end{align*}Where the last line is derived by integrating the integral by parts twice. Since we have $$\mathfrak I'(a,b)$$, we integrate it back with respect to $$b$$ to get$$\mathfrak{I}(a,b)=\arctan(1+b)+C$$To find $$C$$, we observe that when $$a=b$$, then $$\mathfrak{I}(a,b)=0$$. Hence, we get that $$C=\arctan(1+a)$$ so the final answer becomes$$\int\limits_0^1\mathrm dx\,\frac {x^b-x^a}{\log x}\sin\left(\log\frac 1x\right)\color{blue}{=\arctan(1+b)-\arctan(1+a)}$$
• How do you make the transformation? Shoudn't the integer after transformation $x->log\frac1x$ become $\int_0^\infty{sinx}*e^{x(1-b)}dx$. I used the substitution $log\frac1x=u$ , $e^{-u}=x$ then $-\frac1xdx=du$. Don't know where I'm wrong. Can you tell me please how to solve $\int_0^\infty{sinx}*e^{x(1+b)}dx$? And thanks for the help. – David Dan Jan 11 at 18:30
First, under $$x\to e^{-x}$$ $$\frac{\partial I(a,b)}{\partial b}=\int_0^1\sin(\ln(\frac1x))x^bdx=\int_0^\infty e^{(b+1)x}\sin xdx=\frac{1}{(b+1)^2+1}$$ and hence $$I(a,b)=\int_a^b\frac{1}{(s+1)^2+1}ds=\arctan (b+1)-\arctan(a+1).$$