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Parameter dependent integral

parameter-dependent-integral

Parameter-dependent integral is studied in this article. We propose a selection of exercises on continuity, limit, and differentiability of such parametric integrals. Some known transformations in mathematics such as Fourier transform and Laplace transform are parametric integrals.

What is a parametric integral? 

Suppose we have a function of two variables $f:I\times J\to\mathbb{R}$ where $I$ and $J$ are intervals in $\mathbb{R}$. We prefer integrate $f$ with respect to one variable, says the second, we then obtain a function $F:J\to \mathbb{R}$ defined by \begin{align*} F(x)=\int_I f(x,t)dt.\end{align*} We say that the function $F$ is a parameter-dependent integral.

In general, it is very difficult to determine the explicit expression of the function $F$ by simply calculating the integral. But there are theorems that impose conditions on $f$ for which the function $F$ is continuous or differentiable.

Exercises on parameter-dependent integral

Exercise: Let the function \begin{align*} F(x)=\int^{+\infty}_0 \frac{e^{-xt}}{1+t}dt,\qquad \forall x>0. \end{align*}

  • Prove that $F$ is continuous on $(0,+\infty)$.
  • Determine the limit of $F(x)$ as $x\to +\infty$.

Solution: First of all, for any $x>0$, the quantity $F(x)$ is well defined, since \begin{align*} \forall t\ge 0,\quad 0<\frac{e^{-xt}}{1+t}\le e^{-xt},\quad\text{and}\; \int^{+\infty}_0 e^{-xt}dt=\frac{1}{x}. \end{align*}

1) Consider the function \begin{align*} f(x,t)=\frac{e^{-xt}}{1+t},\qquad (x,t)\in (0,+\infty)\times [0,+\infty). \end{align*} Remark that for each fixed $x>0,$ the function $t\in [0,+\infty)\mapsto f(x,t)$ is continuous as the quotient of continuous functions. On the other hand, as the exponential function is continuous, then for each $t\ge 0$ the function $x\in (0,+\infty)\mapsto f(x,t)$ is continuous. It remains to show that on each $[a,b]\times [0,+\infty),$ with $0 < a< b <+\infty$ are arbitrary, the function $f$ is estimated by an integrable function $\varphi [0,+\infty)\to [0,+\infty)$ independent of $x$. In fact for $x\in [a,b]$ and $t\ge 0,$ we have $-x t\le -a t,$ which implies that \begin{align*} |f(x,t)|\le \frac{e^{-at}}{1+t}=:\varphi(t) \end{align*} for all $(x,t)\in [a,b]\times [0,+\infty)$. As for all $t\ge 0,$ $0< \varphi(t)\le e^{-a t}$, it follows that $\varphi$ is integrable. According to the theorem of continuity of parameter integrals, we deduce that the function $F$ is continuous on $(0,+\infty)$.

2) According to the question $1,$ we have proved that the function $f$ is dominated by an integral function $\varphi$. On the other hand, for any $t\ge 0,$ have we have \begin{align*} \lim_{x\to +\infty} f(x,t)=0. \end{align*} By using the dominated convergence theorem, we deduce that \begin{align*} \lim_{x\to +\infty}F(x)=\int^{+\infty}_0 0 dt=0. \end{align*}

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