# Can I apply Dominated convergence theorem for continuous parameter, not natural number?

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 ?

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

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 $$0.$$