# Why is $\int_{0}^{\infty} \frac{1}{1 + x^t} \mathop{\mathrm{d}x}=\frac{\pi/t}{\sin(\pi/t)}$? [duplicate]

I don't know how to show $$\int_{0}^{\infty} \frac{1}{1 + x^t} \mathop{\mathrm{d}x}=\frac{\pi/t}{\sin(\pi/t)}$$ for $t > 1$. I guess I'm supposed to use properties of gamma function and Euler's reflection formula. (I'm sorry if it turns out to be trivial -- I'm not a math student.)

Changing the varible $u=\frac{1}{1+x^t}$ and then using the Beta function, one has \begin{eqnarray} &&\int_{0}^{\infty} \frac{1}{1 + x^t} \mathop{\mathrm{d}x}\\ &=&\frac1t\int_0^1u^{(1-\frac1t)-1}(1-u)^{\frac1t-1}du\\ &=&\frac1tB(1-\frac1t,\frac1t)\\ &=&\frac1t\frac{\Gamma(1-\frac1t)\Gamma(\frac1t)}{\Gamma(1)}\\ &=&\frac1t\frac\pi{\sin(\frac{\pi}{t})}\\ &=&\frac{\pi/t}{\sin(\pi/t)}. \end{eqnarray}