# Finding asymptotic behaviour of integral

I am trying to work out the following integral :

$$\int_0^{at} \mathrm{d}u \frac{e^{-u}}{(t-\frac{u}{a})^\beta}$$

Where $\beta$ is some exponent that I'm trying to evaluate by comparing integrals of that type (for the function to be integrable, I'd expect $\beta<1$).

Now I expect that this integral will be $\sim t^{-\beta}$ for large $t$ :

$$\int_0^{at} \mathrm{d}u \frac{e^{-u}}{(t-\frac{u}{a})^\beta}\\ = \frac{1}{t^\beta}\int_0^{at} \mathrm{d}u \frac{e^{-u}}{(1-\frac{u}{at})^\beta}$$

I am tempted to write the fraction as a series, but I'm blocked because of what happens when $u=at$. Would there be any smart way of getting the asymptotics of this integral ?

Thanks

## 3 Answers

By setting $u=atz$, such that $du=at\,dz$, the given integral equals

$$I(a,\beta,t)=\frac{a}{t^{\beta-1}} \int_{0}^{1}\frac{e^{-at z}}{(1-z)^\beta}\,dz = \frac{a e^{-at}}{t^{\beta-1}}\int_{0}^{1}\frac{e^{at z}}{z^{\beta}}\,dz$$ hence assuming $\beta<1$ we have

$$I(a,\beta,t) = \frac{a e^{-at}}{t^{\beta-1}}\sum_{n\geq 0}\frac{a^n t^n }{n!(n+1-\beta)}$$ by simply expanding the exponential function as its Taylor series and performing termwise integration. Have a look at the Wikipedia page about the incomplete $\Gamma$ function.

• The sum isn't an asymptotic series for large $t$ though, and it doesn't give the leading term. – Maxim Jun 6 '18 at 19:07

You can, in fact, expand the non-exponential part into a series. Or, putting it another way, $\beta$ is fixed, the singularity at $u = a t$ is integrable, and the main contribution still comes from a small neighborhood of $u = 0$: $$\int_0^{a t} \frac {e^{-u}} {(t - u/a)^\beta} du = t \int_0^a \frac {e^{-t \xi}} {(t - t \xi/a)^\beta} d\xi \sim \frac t {(t - t \xi/a)^\beta} \bigg\rvert_{\xi = 0} \int_0^\infty e^{-t \xi} d\xi = t^{-\beta}.$$

As long as $\Re \beta < 1$, the singularity at $u = at$ will not cause you any problem.

Changing variable to $u = ats$, we have

$$\mathcal{I} \stackrel{def}{=} \int_0^{at} \frac{e^{-u}}{(t-\frac{u}{a})^\beta} du = \frac{at}{t^\beta}\int_0^1 \frac{e^{-ats}}{(1-s)^\beta} ds\tag{*1}$$ Notice in the neighborhood of $s=0$, $\frac{1}{(1-s)^\beta}$ is infinitely differentiable with expansion

$$\frac{1}{(1-s)^\beta} = \sum_{k=0}^\infty \frac{(\beta)_k}{k!} s^k\quad\text{ where }\quad (\beta)_k = \prod_{\ell=0}^{k-1} (\beta+\ell) \tag{*2}$$

Furthermore, when $\Re\beta < 1$, we have $$\int_0^1\frac{ds}{|(1-s)^\beta|} = \frac{1}{1 - \Re\beta} < \infty$$

These two observations together allow us to apply Watson's Lemma to the integarl on RHS of $(*1)$. We can read off the asymptotic expansion for $(*1)$ from the expansion in $(*2)$. The end result is

$$\mathcal{I} \asymp \frac{at}{t^\beta}\sum_{k=0}^\infty \frac{(\beta)_k}{k!}\frac{\Gamma(k+1)}{(at)^{k+1}} = \frac{1}{t^\beta}\sum_{k=0}^\infty \frac{(\beta)_k}{(at)^k}$$