$$\int_{0}^{\infty}\dfrac{x^{n-1}}{1+x}dx=\dfrac{\pi}{\sin{n\pi}},\text{ where }0<n<1$$
I want to get the result on right. I substituted $x^{n-1}=t$ but when forwarded to transform in $t.$ The integral gets more complex. I do not know, how to proceed to get the desired result written on the right of the improper integral.
Let $0<n<1$, $f(z)=\frac{z^{n-1}}{(z+1)}$ and $I= ∫_0^{∞}f(x)dx$. Consider the integral of $f(z)$ around the famous contour [given here] (https://en.wikipedia.org/wiki/Contour_integration#/media/File:Keyhole_contour.svg)
Then, $$∫_{C}f(z)dz=∫_{ε}^{M}f(x)dx+∫_{C_{R}}f(z)dz+∫_{M}^{ε}f(x)dx+∫_{C_{ε}}f(z)dz$$ which is equal to $2πi$ times residue of $f(z)$ at $z=-1$, that is, to $2πie^{2π(n-1)i}$. It can easily be shown that the integrals around circles tend to zero as $R→∞$ and $ε→0$. Hence, in the limit we have $(1+e^{2π(n-1)i})I=2πie^{2π(n-1)i}$, that is, $$I=\frac{-2πi}{e^{π(n-1)i}-e^{-π(n-1)i}}=\frac{-π}{\sin(πn-π)}.$$ In the end, $I=\frac{\pi}{\sin πn}.$
We consider two branches of the logarithm. One has a branch cut along the negative imaginary axis where arguments range from $-\frac{\pi}{2}$ to $\frac{3\pi}{2}$. The other has a branch cut along the positive imaginary axis where arguments range from $\frac{\pi}{2}$ to $\frac{5\pi}{2}$. Thus the two branches agree on the half plane $\operatorname{Re} x <0$. Now consider the following contours :
Integrate $\frac{x^{n-1}}{1+x}$ on both contours, with the appropriate branch of $\log x$ then sum them up. After some simplifications (due to the fact that the two branches of $\log x$ agree on the left half plane) we get
$$ \int_{|x+1|=\delta} \frac{x^{n-1}}{1+x}\,\mathrm dx = \int_{|x|=R}\frac{x^{n-1}}{1+x}\,\mathrm dx - \int_{|x|=\delta} \frac{x^{n-1}}{1+x}\,\mathrm dx + \int_\delta^R \big(1- e^{2\pi i(n-1)}\big)\frac{x^{n-1}}{1+x}\,\mathrm dx$$ In the very last integral, we use the regular branch of the logarithm. Residue theorem says that $$ \int_{|x+1|=\delta} \frac{x^{n-1}}{1+x}\,\mathrm dx = 2\pi i e^{\pi i(n-1)} $$ The two remaining integrals can be estimated to be roughly $\delta^{n}$ and $R^{n-1}$ respectively (using the triangle inequality). Since $0<n<1$, these remaining integrals vanish as $\delta \to 0$ and $R\to\infty$, and at the end of the day, we have $$ 2\pi i e^{\pi i(n-1)} = \big(1-e^{2\pi i(n-1)}\big) \int_0^\infty \frac{x^{n-1}}{1+x}\,\mathrm dx $$ Equivalently... $$ \int_0^\infty \frac{x^{n-1}}{1+x}\,\mathrm dx = \frac{\pi}{\sin \pi n} $$
You should use a Gamma-Function. Especially that
$$\Gamma(1-z)\Gamma(z)= \frac{\pi}{\sin\pi z}$$
