I want to calculate $$\lim_{t\to \infty}\int_0^1 \frac{e^{-tx}}{1+x^2}\ dx.$$

My idea is using the Dominated Convergence Theorem (DCT).

Now, for $x\in [0,1]$, $$\displaystyle\left|\frac{e^{-tx}}{1+x^2}\right|\leqq \frac{1}{1+x^2}$$ and RHS is independent of $t$ and integrable on $[0,1]$.

Thus, if I use DCT, I get $$\lim_{t\to \infty}\int_0^1 \frac{e^{-tx}}{1+x^2}\ dx=\int_0^1 \lim_{t\to \infty}\frac{e^{-tx}}{1+x^2}\ dx=0.$$

But I wonder if I'm able to use DCT.

The outline of DCT is :

If $\{f_n\}_{n=1}^\infty$ is a sequence of measurable functions and $\forall x ; f_n(x)\to f(x)$, and there exists integrable $g(x)$ s.t. $|f_n(x)|\leqq g(x) \ \forall n$, then $\displaystyle\lim_{n\to \infty} \int f_n(x)dx=\int f(x)dx.$

In this case, the variable $t$ is not natural number, so I'm not sure whether I can do $\lim_{t\to \infty}\int f(t,x)dx=\int \lim_{t\to \infty}f(t,x) dx.$

Could you explain for this problem ?

  • 2
    $\begingroup$ You can because of the sequential characterization of limits. $\endgroup$
    – Mason
    Jul 28, 2022 at 13:56
  • $\begingroup$ Yes, you can use $\endgroup$
    – MathFail
    Jul 28, 2022 at 14:10
  • $\begingroup$ You should be able to pick any sequence $t_n$ that goes to infinity and apply dominated convergence to that. $\endgroup$ Jul 28, 2022 at 14:14

3 Answers 3


In your case, the dominated convergence theorem applies because for any function $f:[0,\infty)\to\mathbb R$ you have $\lim\limits_{t\to\infty}f(t)=c$ if and only if $\lim\limits_{n\to\infty} f(t_n)=c$ for every sequence $t_n\to\infty$.

For general nets however, DCT may fail. Let $I$ be the irected set of all finite subsets $E\subseteq [0,1]$ and $f_E$ the indicator function of $E$. The net converges pointwise to the indicator function $f$ of $[0,1]$, it is majorized by this $f$ but the integrals are $\int f_E(x)dx=0$ and $\int f(x)dx=1$.


I think you can. Use the following trick. It is true that $[t]\leq\,t$ where [.] the integer part. Then $-t\leq-[t]$ and $-tx\leq\,-[t]x$.

Therefore, $\dfrac{e^{-tx}}{1+x^{2}}$$\leq \dfrac{e^{-[t]x}}{1+x^{2}}$

and the limit you have

$\displaystyle \lim_{t \to +\infty}\int_{0}^{1}\dfrac{e^{-tx}}{1+x^{2}}\leq\displaystyle \lim_{t \to+\infty}\int_{0}^{1}\dfrac{e^{-[t]x}}{1+x^{2}}=\displaystyle \lim_{n \to +\infty}\int_{0}^{1}\dfrac{e^{-nx}}{1+x^{2}}=0$.

Since the integrand is always non-negative, we obtain the required result!


You can use the dominated convergence or Beppo-Levi theorem. However a direct way is the simplest in my opinion.

For $t>0$ we have $$0<\int\limits_0^1{e^{-tx}\over 1+x^2}\,dx\le \int\limits_0^1 e^{-tx}\,dx ={1\over t}(1-e^{-t})\le {1\over t}$$ Thus the limit when $t\to \infty$ is equal $0.$


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