Asymptotic expansion of $I(x)=\int_0^{\infty} \frac{e^{-xt}}{1+t^2}dt$ as $x\rightarrow \infty $

Question

I wish to find the asymptotic expansion of: $$I(x)=\int_0^{\infty} \frac{e^{-xt}}{1+t^2}dt$$ as $x\rightarrow \infty $.

I started by "proving" that: $$ \frac{1}{1+t^2}= \sum_{n=0}^{N}(-1)^nt^{2n} + R_n(t)$$ (I am not allowed to use Watson's lemma :-), which would make it easier).

But how to continue from here?

Answer 1

After a lot of thinking and trying I got the following result: For abbreviation: $f(t) = \frac {1}{1+t^2}$ $$\int_{0}^{\infty} e^{-xt} f(t) dt = \frac {e^{-xt}}{-x} \bigg|_{0}^{\infty} + \frac{1}{x} \int_{0}^{\infty} e^{-xt}f'(t) dt$$ $$\int_{0}^{\infty} e^{-xt} f(t) dt = \frac {1}{x}+\frac {1}{x^2}f'(0) + \frac {1}{x^3}\int_{0}^{\infty} e^{-xt}f'''(t) dt$$ $$\int_{0}^{\infty} e^{-xt} f(t) dt = \frac {1}{x}- \frac {2!}{x^3} +\frac {1}{x^4}f'''(0) + \frac {1}{x^4}\int_{0}^{\infty} e^{-xt}f^{4}(t) dt$$ $$\int_{0}^{\infty} e^{-xt} f(t) dt = \frac {1}{x}- \frac {2!}{x^3} +\frac {4!}{x^5} -\cdots + \frac {1}{x^{2n-1}}f^{2n-1}(0) + \frac{1}{x^{2n}} \int_{0}^{\infty} e^{-xt}f^{2n}(t) dt$$

Answer 2

The simplest way to get an asymptotic approximation to this integral is by integration by parts. So, $$ \int_0^\infty \frac{1}{x}de^{-xt}\frac{1}{1+t^2}=\frac{1}{x}+\frac{2}{x}\int_0^\infty \frac{t}{(1+t^2)^2}e^{-xt}dt $$ and so on.

Your approach is correct but you have to justify the interchange between summation and integration. If you can do it, the result is the following $$ \int_0^\infty e^{-xt}\sum_{n=0}^\infty (-1)^nt^{2n}dt\stackrel{?}{=}\sum_{n=0}^\infty(-1)^n\frac{1}{x^{2n+1}}\Gamma(2n+1). $$ This series is just an asymptotic one.