Prove that (for $n>1$) \begin{equation} \int_{0}^{+\infty} \sin(x^n)dx=\sin\left(\frac{\pi}{2n}\right)\Gamma\left(\frac{n+1}{n}\right). \end{equation} I tried to prove this using a similar argument for Fresnel's integral, but I don't see what I'm doing wrong. Define $\gamma_1:[0,R]\rightarrow\mathbb{C}, \gamma_1(t)=t$, $\gamma_2:[0,\pi/(2n)]\rightarrow\mathbb{C}, \gamma_2(t)=Re^{it}$ and $\gamma_3:[R, 0]\rightarrow\mathbb{C}, \gamma_3(t)=te^{i\pi/(2n)}.$
For Local Theorem of Cauchy, if $f(z)=e^{iz^n}$, then $$\int_{\gamma_1+\gamma_2+\gamma_3} f=0.$$ Now, studying each integral alone, we have for $\gamma_1$, $$\int_{\gamma_1}f=\int_{0}^{R}e^{it^n}dt\rightarrow\int_{0}^{+\infty}\cos(t^n)dt+i\int_{0}^{+\infty}\sin(t^n)dt.$$ For $\gamma_2$, $$\int_{\gamma_2}f=\int_{0}^{\pi/(2n)}e^{i(Re^{it})^n} Rie^{it}dt=Ri\int_{0}^{\pi/(2n)}e^{iR^n(cos(nt)+isin(nt))}e^{it}dt=Ri\int_{0}^{\pi/(2n)}e^{-tR^ncos(nt)}e^{-tiR^nsin(nt))}dt,$$ second exponencial, inside integral, have modulus 1, so if R goes to infinity, integral over $\gamma_2$ is zero because $Re^{-tR^ncos(nt)}\rightarrow0$.
For $\gamma_3$, $$\int_{\gamma_3}f=\int_{R}^{0}e^{i(te^{i\pi/(2n)})^n}e^{i\pi/(2n)}dt=-e^{i\pi/(2n)}\int_{0}^{R}e^{it^ne^{i\pi/2}}dt=-e^{i\pi/(2n)}\int_{0}^{R}e^{-t^n}dt.$$ Then taking imaginary part and R going to infinity, $$\int_{0}^{+\infty} \sin(t^n)dt=\sin\left(\frac{\pi}{2n}\right)\int_{0}^{+\infty} e^{-t^n}dt.$$ Where is my mistake? Any help? Thanks.
You are doing everything correctly. In the end, you only need to do a substitution: Let $y = t^n$, then $dy = nt^{n-1}d t$, or rather $$dt = \frac{1}{n t^{n-1}}dy = \frac{1}{n} y^{-\frac{n-1}{n}}dy$$ and thus (using substitution from $t$ to $y$): \begin{align}\int\limits_0^\infty e^{-t^n}dt &= \frac{1}{n}\int\limits_0^\infty y^{-\frac{n-1}{n}}e^{-y}dy\\ &=\frac{1}{n}\int\limits_0^\infty y^{\left(-\frac{n-1}{n}+1\right)-1}e^{-y}dy\\ &=\frac{1}{n}\Gamma\left(\frac{1}{n}\right)\\ &= \Gamma\left(\frac{1}{n}+1\right)\\ &=\Gamma\left(\frac{n+1}{n}\right)\end{align} Here, we used the definition of the $\Gamma$-function as $$\Gamma(z)=\int\limits_0^\infty t^{z-1}e^{-z}dz$$ and also the identity $z\cdot\Gamma(z)=\Gamma(z+1)$.