Using Integration by Parts to find leading order approximation of exponential integral

Question

I'm trying to understand how finding leading order approximation of exponential integrals.

Here is my integral: $$\int^{+\infty}_0 e^{-xt}\ln(1+t^2)dt$$

I need to use Integration by parts to then find the leading order asymptotic approximation and using the remainder in integral form, showing that the approximation is in fact asymptotic.

(EDIT) This is what I have tried:

Integration by Parts gives:

$$[\frac{e^{-xt}\ln(1+t^2)}{-x}]^{+\infty}_0 + x^{-1}\int^{+\infty}_0e^{-xt}\frac{2t}{1+t^2}dt$$

The boundary term vanishes and the Remainder is the second term.

Then I show that the Remainder is Asymptotic: $$|R(x)| \leq x^{-1}\int^{+\infty}_0|e^{-xt}||\frac{2t}{1+t^2}|dt \leq x^{-1}[\ln(1+t^2)]^{+\infty}_0$$

But I find a divergence.

Any ideas?

Many thanks!

Answer 1

You correctly derived that $$ f(x) = \int_0^\infty e^{-xt} \ln(1+t^2) \, dt = \frac 2x \int_0^\infty e^{-xt} \frac{t}{1+t^2} \, dt $$ for $x > 0$. This does not yet reveal the asymptotic behavior for $x \to \infty$, so let's integrate by parts again: $$ f(x) = -\frac{2}{x^2} e^{-xt} \frac{t}{1+t^2} \big|_{t_0}^{t=\infty} + \frac{2}{x^2} \int_0^\infty e^{-xt} \frac{1-t^2}{(1+t^2)^2} \, dt \, . $$ Again the first part vanishes, and we have $$ f(x) = \frac{2}{x^2} \int_0^\infty e^{-xt} \frac{1-t^2}{(1+t^2)^2} \, dt \, . $$ But we don't give up and integrate once more: $$ \begin{align} f(x) &= -\frac{2}{x^3} e^{-xt} \frac{1-t^2}{(1+t^2)^2} \big|_{t_0}^{t=\infty} + \frac{2}{x^3} \int_0^\infty e^{-xt} \frac{2t(t^2-3)}{(1+t^2)^3} \, dt \\ &= \frac{2}{x^3} + \frac{4}{x^3} \int_0^\infty e^{-xt} \frac{t(t^2-3)}{(1+t^2)^3} \, dt \\ &=: \frac{2}{x^3} + R(x) \, . \end{align} $$ That looks promising! $\frac{t(t^2-3)}{(1+t^2)^3}$ is bounded on $\Bbb R$ (the denominator has no zeros and a higher degree than the numerator). Therefore, with some constant $M > 0$ $$ |R(x)| \le \frac{4M}{x^3} \int_0^\infty e^{-xt} \, dt = \frac{4M}{x^4} $$ and we have shown that $$ f(x) = \frac{2}{x^3} + O\left( \frac{1}{x^4} \right) $$ for $x \to \infty$.

Answer 2

For the antiderivative, using integration by parts, we have $$\int e^{-t x} \log \left(t^2+1\right)\,dx=-\frac 1x e^{-t x} \log \left(t^2+1\right)+\frac 2x\int \frac{t }{t^2+1}\,e^{-t x}\,dt$$ Using the bounds, the first term is $0$ and what remains is the calculation of $$\int_0^\infty \frac{t }{t^2+1}\,e^{-t x}\,dt=\int_0^\infty \frac{t }{(t+i)(t-i)}\,e^{-t x}\,dt$$

Use partial fraction decomposition and then two obvious changes of variable; the integral is finite.

Answer 3

It's been shown that $I\sim2/x^3$ which is the leading order approximation, but for your interest, a complete asymptotic expansion can also be found. $$\ln(1+t^2)=\sum_{n=0}^\infty\frac{(-1)^n}{n+1}t^{2n+2}$$ By Watson's Lemma, $$\boxed{\int_0^{+\infty}\ln(1+t^2)e^{-xt}\ dt\sim\sum_{n=0}^{\infty}(-1)^n\frac{\Gamma(2n+3)}{n+1}\frac{1}{x^{2n+3}}}$$ as $x\to+\infty$.