# Asymptotic expansion of integrals and solving using integration by parts.

I am working on a type of asymptotic expansion

We have $\int_0 ^1 e^{-\frac{x}{t}}dt$ as$x \rightarrow 0+$. In class we have only talked about using integration by parts for integrals of the form $\int f(t) e^{-t}dt$ and Laplace type of integrals. This does not fit the line for either. What I was thinking was to introduce a substitution $u= \frac{1}{t}$, but the problem is that

$$\int_{1} ^\infty \frac{1}{u^2}e^{-xu}du$$ I did this because it now looks like a laplace type of integral. The problem that I see and I think I cannot do much with this form is because for Laplace type of integrals we need $x \rightarrow \infty$ and sadly we do not have that. So I think this is some clever by parts, but I cannot see it. That does not seem correct can someone please help me with this problem.

• What is the problem? What variable do you want an asymptotic expansion in, and in what regime? – Qiaochu Yuan Mar 1 '17 at 4:36
• I just made a small edit as $x$ goes to zero. Sorry I am not familiar what regime is. This is a new topic to me. I believe I need to write this problem in terms of a series of $x$. – Kori Mar 1 '17 at 4:39
• I mean e.g. "as $x \to 0^{+}$" as opposed to "as $x \to \infty$." – Qiaochu Yuan Mar 1 '17 at 4:57
• wolframalpha.com/input/… – tired Mar 1 '17 at 8:03
• integration by parts doesn't help? do this one time and afterwards write $\int_1^{\infty}=\int_0^{\infty}-\int_0^{1}$ and use the integral rep of the euler constant – tired Mar 1 '17 at 8:07

## 2 Answers

Using the change of variables $y = xu$ and integrating by parts we get

$$\int_{1} ^\infty \frac{e^{-xu}}{u^2} \,du = x\int_{x} ^\infty \frac{e^{-y}}{y^2} \,dy \\ = x \left(\frac{e^{-x}}{x} - \int_x^\infty \frac{e^{-y}}{y} \, dy\right)$$

Note that the exponential integral inside the parentheses has a log singularity at $x = 0$ which can be extracted as follows:

\begin{align}\int_x^\infty \frac{e^{-y}}{y} \, dy &= \int_1^\infty \frac{e^{-y}}{y} \, dy - \int_1^x \frac{e^{-y}}{y} \, dy \\ &= \int_1^\infty \frac{e^{-y}}{y} \, dy - \int_1^x \frac{e^{-y}-1}{y} \, dy -\int_1^x \frac{1}{y} \, dy \\ &= \underbrace{\int_1^\infty \frac{e^{-y}}{y} \, dy + \int_0^1 \frac{e^{-y}-1}{y} \, dy}_C -\int_0^x \frac{e^{-y} -1 }{y} \, dy - \log x \\ &= C - \log x - \int_0^x \sum_{n=1}^\infty (-1)^n \frac{y^{n-1}}{n!} \, dy \\ &= C - \log x - \sum_{n=1}^\infty \frac{(-x)^n}{n n!} \end{align}

Thus, we get the following expansion about $x = 0:$

\begin{align}\int_{1} ^\infty \frac{e^{-xu}}{u^2} \,du &= e^{-x} - Cx + x\log x + \sum_{n=1}^\infty \frac{(-x)^{n+1}}{n n!} \\ &= \sum_{n=0}^\infty \frac{(-x)^{n}}{n!} - Cx + x\log x + \sum_{n=1}^\infty \frac{(-x)^{n+1}}{n n!} \\ &= 1 -Cx +x \log x + \sum_{n=1}^\infty \frac{(-1)^{n}x^n (n- x)}{n n!} \end{align}

where $-C$ turns out to be the Euler-Mascheroni constant and the infinite series converges for all $x$, albeit very slowly as $x$ gets large. A good asymptotic approximation is provided with very few terms when $x$ is small.

We immediately see that the integral is asymptotically $\sim e^{-x} \sim 1$ as $x \to 0.$

• this is a nice answer! but to make it also a correct one you have to take into account the series expansion of $e^{-x}=1+\sum_{n\geq1}(-x)^n/n!$ in the end...(the integral is asymtotically $\sim1$ which is clear from your first equation) – tired Mar 1 '17 at 18:12
• @tired: Thanks. I'll add that in. – RRL Mar 1 '17 at 18:18
• nice (+1) from me! – tired Mar 1 '17 at 18:57
• Solid work as always! The final form here has a series with mixed powers $nx^n-x^{n+1}$. I've posted a solution that is similar, but slightly different from yours and wrote the final form as a Taylor series. -Mark – Mark Viola Mar 16 '17 at 20:09

Here, we present an approach that is similar to the one posted by @RRL. However, we write the final form with a Taylor series.

Let $I(x)$ be given by the integral

$$I(x)=\int_1^\infty \frac{e^{-xt}}{t^2}\,dt \tag1$$

Integrating by parts the integral on the right-hand side of $(1)$ with $u=e^{-xt}$ and $v=-\frac1t$ reveals

\begin{align} I(x)&=e^{-x}-x\int_1^\infty \frac{e^{-xt}}{t}\,dt\\\\ &=e^{-x}-x\int_x^\infty \frac{e^{-t}}{t}\,dt\tag2 \end{align}

Integrating by parts the integral on the right-hand side of $(2)$ with $u=e^{-t}$ and $v=\log(t)$, we obtain

\begin{align} I(x)&=e^{-x}+x\log(x)e^{-x}-x\int_x^\infty\log(t)e^{-t}\,dt\\\\ &=e^{-x}+x\log(x)e^{-x}-x\underbrace{\int_0^\infty\log(t)e^{-t}\,dt}_{=-\gamma}+x\int_0^x\log(t)e^{-t}\,dt\\\\ &=e^{-x}+x\log(x)e^{-x}-x\gamma+x\int_0^x\log(t)e^{-t}\,dt\tag3 \end{align}

Using the series representation of $e^{-x}=\sum_{n=0}^\infty\frac{(-1)^nx^n}{n!}$ in $(3)$, interchanging the integral with the series and evaluating the terms $\int_0^x t^n\log(t)\,dt$ by integrating by parts with $u=\log(t)$ and $v=\frac{t^{n+1}}{n+1}$ yields

\begin{align} I(x)&=e^{-x}+x\log(x)e^{-x}-x\gamma+x\sum_{n=0}^\infty\frac{(-1)^n}{n!}\int_0^x\log(t)t^n\,dt\\\\ &=e^{-x}+x\log(x)e^{-x}-x\gamma+x\sum_{n=0}^\infty\frac{(-1)^nx^{n+1}}{(n+1)!}\left(\log(x)-\frac{1}{n+1}\right)\\\\ &=e^{-x}+x\log(x)e^{-x}-x\gamma-xe^{-x}\log(x)+x\log(x)+x\sum_{n=0}^\infty\frac{(-1)^nx^{n+1}}{(n+1)(n+1)!}\\\\ &=x\log(x)-\gamma x+ \color{blue}{e^{-x}-\sum_{n=1}^\infty \frac{(-1)^nx^{n+1}}{n\,n!}}\\\\ &=x\log(x)-\gamma x+ \color{blue}{1-x+\sum_{n=2}^\infty \frac{(-1)^n(2n-1)}{(n-1)\,n!}\,x^n}\\\\ &=1+x\log(x)-(1+\gamma)x+\sum_{n=2}^\infty \frac{(-1)^n(2n-1)}{(n-1)\,n!}\,x^n \end{align}

Note that while this result is the same as the one posted by RRL, it casts the form in a Taylor series of powers of $x$ rather than a series with mixed power terms $nx^n-x^{n+1}$.

• @rrl And thank you. Much appreciative. -Mark – Mark Viola Mar 17 '17 at 1:55
• Please let me know how I can improve my answer. I really want to give you the best answer I can. -Mark – Mark Viola Mar 21 '17 at 3:37