We propose exercises on functions defined by integrals. We mainly prove the regularity of such a function, like continuity and differentiability. These functions can also be considered parameter-dependent integrals.

In many cases, we deal with a function $f(x,t)$ of two variables and we take the integral of this function with respect to one of its variables, for example, $t$. Then we obtain another function $F(x)$. A natural question appears: does properties of $f$ can also be transferred to $F$? In this article, we give some answers over examples.

## A selection of exercises on functions defined by integrals

Problem: The object of this exercise is to calculate the expression of the function defined by integral \begin{align*} \Phi(x)=\int^x_{\frac{1}{x}} f(t)dt\quad\text{with}\;f(t)=\frac{1}{(t+1)^2(t^2+1)} \end{align*}

• The object of this exercise is to calculate the expression of the function defined by integral \begin{align*} \Phi(x)=\int^x_{\frac{1}{x}} f(t)dt\quad\text{with}\;f(t)=\frac{1}{(t+1)^2(t^2+1)} \end{align*}
• Prove that $\Phi$ is differentiable on $(0,+\infty)$ and that \begin{align*} \Phi'(x)=2 f(x),\qquad \forall x\in (0,+\infty). \end{align*}
• Show that for any $x>0,$ \begin{align*} f(x)=\frac{1}{2}\left(\frac{1}{x+1}+\frac{1}{(x+1)^2}-\frac{x}{x^2+1}\right). \end{align*}
• Deduce the expression of $\Phi(x)$ for any $x\in D_\Phi$ “Hint: observe that $\Phi(1)=0$”.
• Application: Let a real $\alpha\in (0,\frac{\pi}{2})$ and consider the integral \\begin{align*} I(\alpha)=\int_\alpha^{\frac{\pi}{2}-\alpha}\frac{\cos^2(\theta)}{1+\sin(2\theta)}d\theta. \end{align*} By using the change of variables $t=\tan(\theta)$, show that \begin{align*} I(\alpha)=-\Phi(\tan(\alpha)). \end{align*} Deduce the expression of $I(\alpha)$.

Solution: 1) Denote by $D_\Phi$ the domain of definition of $\Phi$, i.e. the set of $\mathbb{R}$ in which $\Phi$ is well defined. Clearly, $0\notin D_\Phi$. On the other hand, we have $f(t)>0$ for any $t\in\mathbb{R}\backslash\{-1\}$, so that $\Phi(x)\neq 0$ for any $x\in D_\Phi$. Hence we should exclude $-1$ from $D_\Phi$, so $\{-1,0\}\nsubseteq D_\Phi$. We remark that $\Phi(x)$ is well defined if $-1$ is not in the interval defined by $x$ and $\frac{1}{x}$. Thus $(0,+\infty)\subset D_\Phi$. Let now $x\in (-\infty,0)\backslash\{-1\}$. We distinct two cases: if $x\in (-1,0)$ then $\frac{1}{x} < -1 < x,$ and thus $-1\in (\frac{1}{x},x)$. This shows $(-1,0)\cap D_\Phi=\emptyset$. If $x\in (-\infty,-1)$ then $x < -1 < \frac{1}{x}$. Hence $(-\infty,-1)\cap D_\Phi=\emptyset$. Finally we have \begin{align*} D_\Phi=(0,+\infty). \end{align*}

2) Observe that for any $x>0$, we have $1\in [x,\frac{1}{x}]$ of $1\in [\frac{1}{x},x]$. Thus, Chasless’s Relation implies that for any $x>0,$ we have \begin{align*} \Phi(x)&=\int^x_1 f(t)dt+\int^1_{\frac{1}{x}}f(t)dt\cr &= \int^x_1 f(t)dt-\int_1^{\frac{1}{x}}f(t)dt. \end{align*} Put \begin{align*} F(x)=\int^x_1 f(t)dt,\quad\forall x>0. \end{align*} The function $F$ is a primitive of the continuous $f$, then $F$ is a $C^1$ class function on $(0,+\infty)$, and $F'(x)=f(x)$ for any $x\in (0,+\infty)$. Using this function, we can write \begin{align*} \Phi(x)=F(x)-F\left(\frac{1}{x}\right),\qquad \forall x>0. \end{align*} Hence $\Phi$ is differentiable on $(0,+\infty)$ as sum and composition of differentiable functions. Moreover, for any $x>0,$ \begin{align*} \Phi'(x)&=F'(x)-\left( F\left(\frac{1}{x}\right) \right)’\;\cr & = f(x)-\left(\frac{1}{x}\right)’; F’\left(\frac{1}{x}\right)\cr &= f(x)+\frac{1}{x^2}f\left(\frac{1}{x}\right)\cr &= f(x)+ \frac{1}{x^2}\frac{1}{\frac{(1+x)^2}{x^2}\times \frac{1+x^2}{x^2}}\cr &= 2f(x). \end{align*}

3) It suffices to make a simple calculation \begin{align*} \frac{1}{x+1}+\frac{1}{(x+1)^2}-\frac{x}{x^2+1}&=\frac{(x+1)(x^2+1)+x^2+1-x(x+1)^2}{(x+1)^2(x^2+1)}\cr &=\frac{2}{(x+1)^2(x^2+1)}\cr &=2f(x). \end{align*}

4) Using the fact $\Phi(1)=0$ and using the questions 2 and 3, we obtain, for any $x>0,$ \begin{align*} \Phi(x)&=\int^x_1 \Phi'(t)dt\cr &=\int^x_1 \frac{dt}{t+1}+\int^x_1\frac{dt}{(t+1)^2}-\int^x_1\frac{t}{t^2+1}dt \cr &= \left[\ln(t+1)\right]^x_1+\left[\frac{-1}{t+1}\right]^x_1-\frac{1}{2}\left[\ln(t^2+1)\right]^x_1\cr &= \ln(x+1)-\frac{1}{1+x}-\frac{1}{2}\ln(x^2+1)+\frac{1}{2}-\frac{ln(2)}{2}\cr &= \ln\left(\frac{x+1}{\sqrt{2(x^2+1)}}\right)-\frac{1}{1+x}+\frac{1}{2}. \end{align*}

5) Let the change of variable $t=\tan(\theta)$. Thus \begin{align*} dt= \frac{\theta}{\cos^2(\theta)}= (1+\tan^2(\theta)) d\theta. \end{align*} Thus \begin{align*} d\theta=\frac{dt}{1+t^2}. \end{align*} As $\tan(\frac{\pi}{2}-\alpha)=\frac{1}{\tan(\alpha)}$, then \begin{align*} I(\alpha)=\int^{\frac{1}{\tan(\alpha)}}_{\tan(\alpha)} \frac{1} {\frac{1+\sin(2\theta)}{\cos^2(\theta)}} \frac{dt}{1+t^2}. \end{align*} Since $\sin(2\theta)=2\sin(\theta) \cos(\theta)$ and $\frac{1}{\cos^2(\theta)}=1+\tan^2(\theta)$, we have \begin{align*} \frac{1} {\frac{1+\sin(2\theta)}{\cos^2(\theta)}}&= \frac{1}{1+\tan^2(\theta)+2\tan(\theta)}\cr &= \frac{1}{(1+\tan(\theta))^2}\cr &= \frac{1}{(1+t)^2}. \end{align*} This implies that \begin{align*} I(\alpha)= \int^{\frac{1}{\tan(\alpha)}}_{\tan(\alpha)} \frac{dt}{(1+t)^2(1+t^2)}=-\Phi(\tan(\alpha)). \end{align*} Finally, \begin{align*} I(\alpha)&= – \ln\left(\frac{\tan(\alpha)+1}{\sqrt{2(\tan^2(\alpha)+1)}}\right)+\frac{1}{1+\tan(\alpha)}-\frac{1}{2}\cr &= – \ln\left(\frac{\sin(alpha)+\cos(alpha)}{\sqrt{2}}\right)+\frac{1}{1+\tan(\alpha)}-\frac{1}{2}. \end{align*}

The reader can also consult a study of some simple examples of such integral functions.