Calculus

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 \(R\) . Note that \(f(z)\) had two singularities, one at \(z=-i\) and the other at \(z=i\) . Only \(z=i\) is inside of our contour \(C\) , as seen below:

Here we will solve the Dirichlet Integral using two different techniques. The Dirichlet Integral is defined as $$ \int_{0}^{\infty} \frac{\sin(x)}{x} dx $$

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. Thus, $$ I=e^{-\frac{b^2}{4a}+c} \int_{-\infty}^{\infty} e^{at^2} dt $$ and since \(a\) is negative, we can write $$ I=e^{-\frac{b^2}{4a}+c} \int_{-\infty}^{\infty} e^{-(\sqrt{-a}t)^2} dt $$ Now letting \(v=\sqrt{-a}t\) we have that \(dv=\sqrt{-a} dt \implies \frac{dv}{\sqrt{-a}}=dt\) and thus $$ I=e^{-\frac{b^2}{4a}+c} \int_{-\infty}^{\infty} e^{-(v)^2} \frac{dv}{\sqrt{-a}} \implies I=\frac{e^{-\frac{b^2}{4a}+c}}{\sqrt{-a}} \sqrt{\pi} = e^{-\frac{b^2}{4a}+c} \sqrt{-\frac{\pi}{a}} $$ and we’re done.

The Gaussian Integral is given by $$ \int_{-\infty}^{\infty} e^{-x^2} dx $$ This is a beautiful integral with many important applications to probability and statistics, namely normal distributions. Here, we will evaluate this integral using techniques from Calculus 3.

We will evalute the integral \(\int_0^1 \frac{\ln(x)}{1+x} dx \) using Taylor Series expansion and switching the order of integration and summation.

Firstly, we notice that $$ \frac{1}{1+x}=\sum_{n=0}^{\infty} (-1)^n x^n $$

by taylor series expansion. This series coverges for \(x \in (0,1)\)

so we can substitute this into the integral:

$$ \int_0^1 \frac{\ln(x)}{1+x} dx = \int_0^1 \ln(x) \sum_{n=0}^{\infty} (-1)^n x^n dx $$

Further, by changing the order of integration and summation, we have $$ \int_0^1 \ln(x) \sum_{n=0}^{\infty} (-1)^n x^n dx = \sum_{n=0}^{\infty} (-1)^n \int_0^1 x^n \ln(x) dx $$