Owen's Math Portal

Today, we will evaluate the integral $$ \int_{-\infty}^{\infty} \frac{\cos(x)}{x^2+1} dx $$ using Cauchy’s Residue Theorem and contour integration. This is one of the most beautiful results in all of calculus, and the solution is rather elegant. To begin, let us define the following function in \(\mathbb{C}\) : $$ f(z)=\frac{e^{iz}}{z^2+1} $$ and let us use the canonical contour, that is, the top half of a semicircle in the complex plane. Let us define this contour as \(C\) , with radius \(R\) .

Here we will solve the Dirichlet Integral using two different techniques. The Dirichlet Integral is defined as $$ \int_{0}^{\infty} \frac{\sin(x)}{x} dx $$ Feynman Integration We employ Feynman’s Integration trick here by first defining the function $$ I(\alpha)=\int_{0}^{\infty} \frac{\sin(x)}{x} e^{-\alpha x} dx $$ and notice that because \(|\frac{\sin(x)}{x}| \leq 1 \) $$ \lim_{\alpha \rightarrow \infty} |\int_{0}^{\infty} \frac{\sin(x)}{x} e^{-\alpha x} dx| \leq \int_{0}^{\infty} e^{-\alpha x} dx = \frac{1}{\alpha} $$ so \(I(\alpha) \) vanishes as \(\alpha \rightarrow \infty \) and thus we can continue with Fenyman’s technique: $$ \frac{\partial I(\alpha)}{\partial \alpha} = \frac{\partial}{\partial \alpha} \int_{0}^{\infty} \frac{\sin(x)}{x} e^{-\alpha x} dx = -\int_{0}^{\infty} x \frac{\sin(x)}{x} e^{-\alpha x} dx = -\int_{0}^{\infty} \sin(x) e^{-\alpha x} dx $$ now by integration by parts, we have that $$ -\int_{0}^{\infty} \sin(x) e^{-\alpha x} dx = -1+\alpha \int_{0}^{\infty} \cos(x) e^{-\alpha x} dx $$

Today, we will be evaluating the integral $$ \int_{-\infty}^{\infty} e^{ax^2+bx+c} dx \hspace{1cm} a \in \mathbb{R^{-}}, b \in \mathbb{R}, c \in \mathbb{R} $$ The restriction of \(a\) to \(\mathbb{R^{-}}\) is required for the convergence of this integral. My thought process for this integral was initially to minipulate the integrand in such a way that we can employ the result of the Gaussian integral, namely, \(\int_{-\infty}^{\infty} e^{-x^2} dx = \sqrt{\pi}\) . To do this, we complete the square on \(f(x)=ax^2+bx+c\) : $$ y=ax^2+bx+c \implies y-c=ax^2+bx=a(x^2+\frac{b}{a}x) \implies \frac{y-c}{a}=x^2+\frac{b}{a}x $$ Further, $$ \frac{y-c}{a}+\frac{b^2}{4a^2}=x^2+\frac{b}{a}x+\frac{b^2}{4a^2} \implies \frac{y-c}{a}=(x+\frac{b}{2a})^2-\frac{b^2}{4a^2} $$ Lastly, $$ y=a(x+\frac{b}{2a})^2-\frac{b^2}{4a}+c $$ Thus we have that $$ I=\int_{-\infty}^{\infty} e^{ax^2+bx+c} dx=\int_{-\infty}^{\infty} e^{a(x+\frac{b}{2a})^2-\frac{b^2}{4a}+c} dx=e^{-\frac{b^2}{4a}+c} \int_{-\infty}^{\infty} e^{a(x+\frac{b}{2a})^2} dx $$ Given that this integrand spans all of \(\mathbb{R}\) we can employ the substitution \(t=x+\frac{b}{2a}\) and our bounds will remain the same.