How should I set the integral route for this complex integral?

Question

\begin{equation} \int_0^{\infty} \dfrac{x^{\lambda -1}}{x^2+1} dx = \dfrac{\pi}{2 \sin{\dfrac{\pi \lambda}{2}}} \quad (0< \lambda < 2) \end{equation}

I will prove this using complex integral, but I don't come up with how I should set the integral route.

I tried prove the equality using the integral route below and using residue theorem.

$ C_{R} : \text{semicircle in first and second quadrant whose radius is R }\\ C_1 : [-R, \epsilon]\\ C_{\epsilon} : \text{semicircle in first and second quadrant whose radius is $\epsilon$ }\\ C_2 : [\epsilon, R] $

However, I failed to prove. It seems that it is impossible to calculate the integral with this integral route.

I would like you to teach me how I should set the integral route.

Answer 1

One may use $\beta$ function.

$$I=\int_{0}^{\infty} \frac{x^{k-1}}{1+x^2} dx$$ Let $x=\tan t$, then $$I=\int_{0}^{\pi/2} \sin^{k-1}t ~\cos^{1-k}t ~dt=\frac{1}{2} \frac{\Gamma(k/2)\Gamma(1-k/2)}{\Gamma(1)}= \frac{\pi}{2}\csc(\pi k/2), 0<k<2$$

See for $\beta$ function: https://en.wikipedia.org/wiki/Beta_function

Answer 2

I will use $a$ instead of $\lambda$, it is simpler to type. Since $a-1$ may take negative values, we may have some problems in $0$, so the contour from the OP avoiding $0$ is needed. The following is valid for $$ a\ne 1\ . $$ (For this particular value use an other argument or a posteriori the continuity in $a$.)

Let us denote the two contour pieces $C_1$ and $C_2$ from the OP by $C_1(\epsilon,R)$ and $C_2(\epsilon,R)$, also involving $\epsilon$ and $R$.

The expression $x^a$ is then rather defined as $\exp(a \log x)$ with a branch $\log$ for the logarithm defined to coincide with the real natural logarithm $\ln$ on $(0,\infty)$. So we have for $x>0$ the relation $$(-x)^{a-1}=(-1)^{a-1}\cdot x^{a-1}=\exp(i\pi\; (a-1))\cdot x^{a-1}\ .$$

So let us compute the integral $$ \begin{aligned} J(a) &:= \int_0^{\infty} \dfrac{x^{a -1}}{x^2+1} \;dx \\ &=\lim_{\substack{0<\epsilon<R\\\epsilon \to0\\R\to\infty}} \int_\epsilon^\infty \dfrac{x^{a -1}}{x^2+1} \;dx =\lim_{\substack{0<\epsilon<R\\\epsilon \to0\\R\to\infty}} \int_{C_2(\epsilon,R)} \dfrac{x^{a -1}}{x^2+1} \;dx\ . \\[3mm] &\qquad\text{We have:} \\[3mm] \int_{-\infty}^0 \dfrac{x^{a -1}}{x^2+1} \;dx & =\int_0^{\infty} \dfrac{(-x)^{a -1}}{x^2+1} \;dx =e^{i\pi \; a}\cdot \int_0^{\infty} \dfrac{(-x)^{a -1}}{x^2+1} \;dx \\ &= e^{i\pi\; a}J(a)\ . \\[3mm] \left|\int_{C(\epsilon)} \dfrac{x^{a -1}}{x^2+1} \;dx \right| &\le \int_{C(\epsilon)} \dfrac{|x|^{a -1}}{|x^2+1|} \;dx \le \int_{C(\epsilon)} \dfrac{\epsilon^{a -1}}{1-\epsilon^2} \;dx =\pi\epsilon\cdot\dfrac{\epsilon^{a -1}}{1-\epsilon^2} \\ &\to 0\qquad\text{ because $a>0$.} \\[3mm] \left|\int_{C(R)} \dfrac{x^{a -1}}{x^2+1} \;dx \right| &\le \int_{C(R)} \dfrac{|x|^{a -1}}{|x^2+1|} \;dx \le \int_{C(\epsilon)} \dfrac{R^{a -1}}{R^2-1} \;dx =\pi R\cdot\dfrac{R^{a -1}}{R^2-1} \\ &\to 0\qquad\text{ because $a<2$.} \\[3mm] &\qquad\text{This gives:} \\[3mm] (1+e^{i\pi\; (a-1)})\cdot J(a) &= \lim_{\substack{0<\epsilon<R\\\epsilon \to0\\R\to\infty}} \int_{C_1(\epsilon,R)\ \sqcup\ C_2(\epsilon,R)} \dfrac{x^{a -1}}{x^2+1} \;dx \\ &= \lim_{\substack{0<\epsilon<R\\\epsilon \to0\\R\to\infty}} \int_{C_1(\epsilon,R)\ \sqcup \ C(\epsilon)\ \sqcup \ C_2(\epsilon,R)\ \sqcup \ C(R)} \dfrac{x^{a -1}}{x^2+1} \;dx \\ &=2\pi\cdot i\sum_{r\text{ Residue}}\operatorname{Res}_{x=r}\dfrac{x^{a -1}}{x^2+1} \\ &=2\pi\cdot i\cdot \operatorname{Res}_{x=i}\dfrac{x^{a -1}}{x^2+1} \\ &=2\pi\cdot i\cdot \operatorname{Res}_{x=i}\frac 1{2i}x^{a -1} \left(\frac 1{x-i}-\frac 1{x+i}\right) \\ &=2\pi\cdot i\cdot \operatorname{Res}_{x=i}\frac 1{2i}x^{a -1}\cdot \frac 1{x-i} \\ &=\pi\cdot i^{a-1}=\pi\cdot e^{i\frac \pi 2(a-1)}\ . \end{aligned} $$ Now $1+e^{ib}=(1+\cos b)+i\sin b=2\cos ^2\frac b2+2i\sin\frac b2\cos \frac b2=2\cos\frac b2\cdot e^{ib/2}$.

This gives finally: $$ J(a) = \frac {\pi\cdot \color{blue}{e^{i\frac \pi 2(a-1)}}} {2\cos\frac \pi 2(a-1)\cdot \color{blue}{e^{i\frac \pi 2(a-1)}}} = \frac \pi {2\cos\frac \pi 2(1-a)} = \frac \pi {2\sin\frac {\pi a} 2} \ . $$ $\square$