# upper bounds for $\int_a^{b} \frac{\exp(x)}{x}\ dx$

Let $$a be a positive real numbers. Are there tight upper bounds for $$\int_a^{b} \frac{\exp(x)}{x}\ dx$$, specially asymptotic bounds when $$a, b,\frac{b}{a}\to\infty$$?

• The given integral is the exponential integral. Sep 3 '19 at 23:05
• What is your progress so far? Sep 3 '19 at 23:27

Applying integration by parts $$n$$ times gives

\begin{align*} \int_{a}^{b} \frac{e^x}{x} \, \mathrm{d}x = e^b R_n(b) - e^a R_n(a) + \int_{a}^{b} \frac{n! e^{x}}{x^{n+1}} \, \mathrm{d}x, \end{align*}

where $$R_n(x) = \sum_{k=1}^{n} \frac{(k-1)!}{x^k}$$. Now, as $$a, b, (b/a) \to \infty$$, the last integral is bounded by $$\mathcal{O}(e^b / b^{n+1})$$, and so, we get

$$\int_{a}^{b} \frac{e^x}{x} \, \mathrm{d}x = e^b R_{n}(b) + \mathcal{O}(e^b /b^{n+1}).$$

as $$a, b, (b/a) \to \infty$$, for each fixed $$n \geq 1$$.

The integral will be asymptotically $$\frac{e^b}b$$.

The lower bound $$\int_a^b \frac{e^x}x\,dx > \int_a^b \frac{e^x}b\,dx = \frac{e^b-e^a}b$$ is easy. For an upper bound, for any $$c\in(a,b)$$ we have \begin{align*} \int_a^b \frac{e^x}x\,dx &= \int_a^c \frac{e^x}x\,dx + \int_c^b \frac{e^x}x\,dx \\ &< \int_a^c \frac{e^x}a\,dx + \int_c^b \frac{e^x}c\,dx \\ &= \frac{e^c-e^a}a + \frac{e^b-e^c}c; \end{align*} taking $$c$$ close (but not too close) to $$b$$, for example $$c=b-\sqrt{ab}$$ (valid when $$\frac ba$$ isn't too small), recovers the asymptotic $$\frac{e^b}b$$.

One takeaway: always try trivial bounds first; and if the integrand (or summand) consists of an increasing function times a decreasing function, try cutting the integral (or series) at an arbitrary middle point and try trivial bounds on each portion separately.