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I need to solve this integral $\displaystyle \int_{0}^{1} \frac{(1-x)^\alpha}{(\ln x)^2 \sin \sqrt{x}} \,dx \,$ for $\alpha > 0$, how can I do?

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    own thoughts $ $? – 2017-01-20
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    Do you mind to share the origin of this beast? It looks like something someone just made up to give a very strange example. – 2017-01-20
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    "simple" exsercise of analysis 1 of university – 2017-01-20
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    i think the question is concerned with the convergence of the integral instead of giving a closed form solution, right? – 2017-01-20
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    can be... lg in 0 is not defined – 2017-01-20
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    @Mazzorca, *you* should know what the excersise asks you to do. Or did someone just give you this intergal and tell you to do whatever you want with it? – 2017-01-20

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There is no way to compute this integral, except maybe for special values of $\alpha$ ...

Il will only try to find the values of $\alpha$ such that this integral is convergent.

In the vincinity of $0$ :

$$f_\alpha(x)=\frac{(1-x)^\alpha}{\ln^2(x)\sin(\sqrt x)}\underset{0}{\sim}\frac{1}{\sqrt{x}\ln^2(x)}=\frac{x^{1/4}}{\ln^2(x)}\,\frac{1}{x^{3/4}}$$Since $\displaystyle{\lim_{x\to0}\frac{x^{1/4}}{\ln^2(x)}=0}$, there exists $\delta\in(0,1)$ such that :

$$\forall x\in(0,\delta],\,0\le f_\alpha(x)\le\frac{1}{x^{3/4}}$$

This proves that $\int_0^{1/2}f_\alpha(x)\,dx$ converges

In the vincinity of 1 :

$$f_\alpha(x)\underset{1}{\sim}\frac{1}{\sin(1)}(1-x)^{\alpha-2}$$

The required condition for the convergenceof $\int_{1/2}^1f_\alpha(x)\,dx$ is therefore $\alpha-2>-1$.

$$\boxed{\left(\int_0^1f_\alpha(x)\,dx\quad\mathrm{converges}\right)\Leftrightarrow(\alpha>1)}$$

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    This is a good contribution, although "no way to compute this integral" is a rather bold statement, after all it can be computed numerically and it might be possible to express it in some other way, i.e. a series or a combination of special functions. – 2017-01-20
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    I agree ... I should have said "I cannot see any way to compute explicitly this integral in terms of usual functions, except maybe for special values of $\alpha$". – 2017-01-20
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    Still, the spirit of your statement is absolutely correct, since Wolfram Alpha can't even evaluate the exact value for $\alpha=2$ – 2017-01-20