We offer a selection of improper integral exercises with detailed answers. Our main objective is to show you how to prove that an improper integral is convergent: how to employ the integration by parts; how to make a change of variables; how to apply the dominated convergence theorem; and how to integrate terms by terms.
Integrals on unbounded intervals, improper integral?
Proper Riemann integrals are defined for bounded functions on bounded intervals. Now we ask the question: can we define integrals for functions that are unbounded in the neighborhood of a point $(a,b]$ or functions defined on unbounded intervals of the form $(-\infty,+\infty),$ $[a,+infty),$ $(-\infty,a].$
Let $f:[a,+\infty)\to\mathbb{R}$ be a continuous function. If the limit $$ \lim_{x\to+\infty}\int_a^x f(t)dt$$ exists, then we say that the integral \begin{align*} \int^{+\infty}_a f(t)dt\end{align*} is convergent, and the value of this integral is exactly the value of the limit.
Similarly we define the improper integral of a continuous function on $(-\infty,a]$.
Let $f:[a,b)\to\mathbb{R}$ be a continuous function. If the limit $$ \lim_{x\to b^-}\int_a^x f(t)dt$$ exists, then we say that the integral \begin{align*} \int^{b}_a f(t)dt\end{align*} is convergent.
Selected improper integral exercises
Exercise: Show that the following integrals are convergent \begin{align*} \begin{array}{cc} 1.\; \displaystyle\int^{+\infty}_0 e^{-t^2}dt & \quad 2.\; \displaystyle\int^{\frac{\pi}{2}}_0 \sqrt{\tan(\theta)};d\theta\\ 3.\;\displaystyle\int^{+\infty}_0 \frac{\ln(t)}{1+t^2}dt & \quad 4.; \displaystyle\int^1_0 \frac{dt}{1-\sqrt{t}}. \end{array} \end{align*}
Solution: 1) It suffices to show the convergence on the interval $[1,+\infty)$. For any $t\ge 1,$ we have $t^2\ge t,$ so that \begin{align*} 0< e^{-t^2}\le e^{-t},\qquad \forall t\ge 1. \end{align*} Observe that \begin{align*} \int^{+\infty}_1 e^{-t}dt&=\lim_{x\to +\infty} \int^x_1 (-e^{-t})’dt\cr &= \lim_{x\to+\infty} (e^{-1}-e^{-x})\cr &= \frac{1}{e}. \end{align*} This implies that the integral \begin{align*} \int^{+\infty}_0 e^{-t}dt \end{align*} is convergent. Hence the integral \begin{align*} \int^{+\infty}_0 e^{-t^2}dt \end{align*} is convergent is convergent as well.
Another proof: Remark that $t^2 e^{-t^2}\to 0$ as $t\to +\infty$. This means that there exists $\gamma>0,$ sufficiently large, such that $t^2 e^{-t^2}<1$ for any $t\ge \gamma$. We know that the integral \begin{align*} \int_\gamma^{+\infty} \frac{dt}{t^2} \end{align*} converge. Then by comparison the integral \begin{align*} \int_\gamma^{+\infty} e^{-t^2}dt \end{align*} converges. This ends the proof.
2) Here we shall use a change of variables. We have problem near of $\frac{\pi}{2},$ as $\cos(\frac{\pi}{2})=0$. We need to define a function of class $C^1$ which will play the role of change of variable. We then define the function \begin{align*} \psi(\theta)=\sqrt{\tan(\theta)},\quad \theta\in \left(0,\frac{\pi}{2}\right). \end{align*} Clearly $\psi$ is a $C^1$ function on $\left(0,\frac{\pi}{2}\right)$. In addition, for any $0< \theta < \frac{\pi}{2}$, we have \begin{align*} \psi'(\theta)&=\frac{\tan'(\theta)}{2 \sqrt{\tan(\theta)}}\cr &= \frac{\tan^2(\theta)}{2 \sqrt{\tan(\theta)}}. \end{align*} We deduce that $\psi'(\theta)>0$ for any $0< \theta < \frac{pi}{2}$, so that $\psi$ define a bijection from $\left(0,\frac{\pi}{2}\right)$ to \begin{align*}\left(\psi(0),\lim_{\sigma\to \frac{\pi}{2}}\psi(\sigma)\right)=(0,+\infty).\end{align*} Let $x\in (0,+\infty)$ and $0< \theta < \frac{\pi}{2}$ such that $x=\psi(\theta)$, which means that $x^2=\tan(\theta)$. Thus $\theta=\arctan(x^2)$. It follows that \begin{align*} d\theta=\frac{2x}{1+x^4} dx. \end{align*} Now we can write \begin{align*} \int^{\frac{\pi}{2}}_0 \sqrt{\tan(\theta)}\;d\theta=2\int^{+\infty}_0 \frac{x^2}{1+x^4}dx \end{align*} The function $f(x)=\frac{x^2}{1+x^4}$ is defined and continuous on $[0,+\infty)$. Observe that \begin{align*} 0 < f(x)\le \frac{1}{x^2},\quad\forall x\ge 1. \end{align*} An the function $x\mapsto \frac{1}{x^2}$ is integrable on $[1,+\infty),$ then the function $f$ is integrable on $[0,+\infty)$. This ends the proof.
3) The function $g(t)=\frac{\ln(t)}{1+t^2}$ is continuous on $(0,+\infty)$. Moreover, \begin{align*} t^{\frac{3}{2}} f(t)\quad\underset{t\to+\infty}{\sim}\quad \frac{\ln(t)}{\sqrt{t}}. \end{align*} As $t^{-\frac{1}{2}}\ln(t)\to 0$ as $t\to +\infty$, then \begin{align*} \lim_{t\to +\infty} t^{\frac{3}{2}} f(t)=0. \end{align*} On the other hand, as the function $t\mapsto \frac{1}{ t^{\frac{3}{2}}}$ is integrable on $[1,+\infty)$, the function $f$ is integrable on $[1,+\infty)$.
Clearly we have \begin{align*} \sqrt{t} f(t)\quad\underset{t\to+\infty}{\sim}\quad \sqrt{t} \ln(t). \end{align*} Thus \begin{align*} \lim_{t\to 0}\sqrt{t} f(t)=0. \end{align*} We know that the function $t\mapsto \frac{1}{\sqrt{t}}$ is integrable on $(0,1]$, it follows that the function $f$ is integrable on $(0,1]$. Conclusion: $f$ is integrable on $(0,+\infty)$. The improper integral is then convergent.
The following problem is classical in improper integral exercises.
Exercise “Gamma function”: For $x\in\mathbb{R},$ we consider the integral \begin{align*} \Gamma(x):=\int^{+\infty}_0 t^{x-1}e^{-t}dt. \end{align*} We also denote $I:=\{x\in \mathbb{R}: \Gamma(x) < \infty\}$.
- Determine $I$.
- Show that for all $x\in I,$ $\Gamma(x+1)=x\Gamma(x)$. Deduce the value of $\Gamma(n)$ for any $n\in\mathbb{N}^\ast$.
- Justify that \begin{align*} \Gamma\left(\frac{1}{2} \right)=2\int^{+\infty}_0 e^{-t^2}dt. \end{align*}
Solution: 1) Let $x\in\mathbb{R}$. the function $f:t\mapsto t^{x-1}e^{-t}$ is continuous on $(0,+\infty),$ as product of continuous functions. On the other hand, clearly we have $t^2f(t)\to 0$ as $t\to +\infty$. This shows that $f(t)$ is equivalent to $\frac{1}{t^2}$ when $t$ is near of $+\infty$. Thus the function $f$ is integrable on $[1,+\infty)$. In addition $f(t)\sim t^{x-1}$, $t\to 0$, so that $f$ is integrable on $(0,1]$ if and only if $x-1>1,$ i.e. $x>0$. Hence $f$ is integrable on $(0,+\infty)$ if and only if $x>0$. Finally $I=(0,+\infty)$.
2) Let $x\in I$. The functions $u:t\mapsto t^x$ and $v:t\mapsto e^{-t}$ are of class $C^1$ on $(0,+\infty)$ and $u(t)v(t)\to 0$ as $t\to 0$ and $t\to +\infty$. Now by integration by parts, we have \begin{align*} \Gamma(x+1)=\left[t^x (-e^{-t})\right]^{+\infty}_0 -\int^{+\infty}_0 xt^{x-1}(-e^{-t})dt=x\Gamma(x). \end{align*} Let now $n\in \mathbb{N}^\ast$. First, observe that \begin{align*} \Gamma(1)=\int^{+\infty}_0 e^{-t}dt= \left[-e^{-t}\right]^{+\infty}_0=1. \end{align*} We then have \begin{align*} \Gamma(n)=n\Gamma(n-1)=n(n-1)\Gamma(n-2)=\cdots=n(n-1)\cdots 2\Gamma(1)=n!. \end{align*}
3) In the integral defining $\Gamma\left(\frac{1}{2} \right),$ we make the change of variables $t=u^2$, and we obtain \begin{align*} \Gamma\left(\frac{1}{2} \right)=\int^{+\infty}_0 t^{-\frac{1}{2}}e^{-1}dt=\int^{+\infty}_0 \frac{e^{-u^2}}{u}2udu=2\int^{+\infty}_0 e^{-u^2}du. \end{align*}
