How to Integrate e^(1/x)
The function y = e^(1/x) is a solution to the differential equation (x^2)y' = -y. It has a horizontal asymptote at y = 1, meaning as x goes to infinity or negative infinity, the function e^(1/x) tends toward the value of 1. However, at x = 0, the behavior of e^(1/x) is a little strange. On the positive side of the x-axis, e^(1/x) goes to infinity as x goes to 0. But on the negative side, e^(1/x) goes to 0 as x goes to 0. See graph below.
A function of the form h(x) = e^(x^n) can be integrated only if the exponent n is equal to 1, 1/2, 1/3, 1/4, 1/5, etc. That means f(x) = e^(x^-1) = e^(1/x) is non-integrable; in other words, you can't calculate the integral of e^(1/x) in terms of elementary functions. It is possible, however, to approximate the antiderivative of e^(1/x) with an infinite series, and use that series to estimate definite integrals. You can also use Riemann sums to estimate definite integrals.
An Infinite Series for the Integral of e^(1/x)
The Taylor series expansion of ew is
ew = 1 + w + w2/2 + w3/6 + w4/24 + w5/120 + ... + wn/n! + ...
If we replace w with 1/x, we get a series for e1/x:
e1/x = 1 + 1/x + 1/(2x2) + 1/(6x3) + 1/(24x4) + 1/(120x5) + ...
Now if we integrate this series term by term we can obtain an infinite series for the integral of e^(1/x):
∫ e1/x dx = x + Ln|x| - 1/(2x) - 1/(12x2) - 1/(72x3) - 1/(480x4) - ...
This series is convergent for all values of x except for 0, so you can truncate it somewhere and use it to estimate the definite integral of e1/x from x = -2 to x = -0.5, or from x = 3 to x = 7, for example. But you can't use it to evaluate the integral from -2 to 3, since that interval includes the discontinuity at 0, and so does not converge. The more terms you include in the truncation, the more accurate the estimate.
An Asymptotic Series for the Integral of e^(1/x)
We can use the technique of integration by parts to find an asymptotic series (non-convergent series) for the integral of e1/x that provides a good estimate for values of x that are near zero. First we rewrite the integral in an equivalent form.
∫ e1/x dx = ∫ [-e1/x / x2] * [-x2] dx
If we let v' = -e1/x/x2 and u = -x2, then we get u' = -2x and v = e1/x. This gives us the new integral equation
∫ e1/x dx = -x2e1/x + 2∫ xe1/x dx
Now we replace the integrand x*e^(1/x) with the equivalent [-e1/x/x2]*[-x3] and apply integration by parts again using the same assignment with v' = -e1/x/x2. This gives us
∫ e1/x dx = -x2e1/x - 2x3e1/x - 6∫ x2e1/x dx
Repeating this process over and over gives us the infinite series
∫ e1/x dx = -[e1/x]*[x2 + 2x3 + 6x4 + 24x5 + ... + n!xn+1 + ...]
More Integration Tutorials
- How to Integrate sin(x)/x and cos(x)/x
- How to Integrate sin(x^2) and cos(x^2)
- How to Integrate e^(x^(1/2)) = e^sqrt(x)
For general help with integration techniques and finding antiderivatives, see also the Table of Integrals and Four Common Mistakes Students Make with u-Substitution.
This series doesn't converge for any value of x, however, if you truncate it after a finite number of terms and plug in a values of x that are near zero and on the same side of the x-axis, then it provides a good estimate of the definite integral. For example, you can use it to evaluate the integral of e1/x from x = 0.1 to x = 0.2 using four terms.
How many terms you should use depends on the limits of the integral. If the limits are x = p to x = q, with p and q positive and q greater than p, then you should use (1/q - 1) terms, where (1/q - 1) is rounded up to a whole number. For example, if q = 0.15, you would use 6 terms, since (1/0.15 - 1) = 5.67.
If p and q are negative with q less than p, use (1/|q| - 1) terms, with (1/|q| - 1) rounded up to a whole number.
One way in which asymptotic series are different from convergent series is that if you include more terms in the truncation, you actually get a worse estimate.
We can use the first series to estimate the definite integral of e1/x from x = -3 to x = -1. This region is shaded in green in the figure below.
Using the truncated series
∫ e1/x dx ≈ x + Ln|x| - 1/(2x) - 1/(12x2) - 1/(72x3) - 1/(480x4)
we plug in x = -1 and x = -3 and subtract. This gives us
[-823/1440] - [Ln(3) - 110501/38880] ≈ 1.171964
With a numerical integrator, we find the area under the curve from x = -3 to x = -1 is closer to 1.172211, so the truncated series with six terms underestimates the area by about 0.000247. With more terms we would get an even more accurate approximation.
Let's use the asymptotic series to estimate the area under the curve y = e1/x from x = 0.12 to x = 0.18. This area is shaded in green in the graph below.
Since the upper endpoint is 0.18, we compute 1/0.18 - 1 = 4.56 and round this up to 5. This means we take 5 terms of the asymptotic series. Our integral approximation formula is then
∫ e1/x dx ≈ [-e1/x]*[x2 + 2x3 + 6x4 + 24x5 + 120x6]
Plugging in the endpoints 0.18 and 0.12 and subtracting gives us
-15.256133 - (-83.436862) = 68.180729
Using a numerical integrator gives us an answer closer to 70.9762, so our estimate is off by about 4% from the true answer. We could get a more accurate estimate using the convergent series in the first example, but we would need many more terms.
Is the Improper Integral of e^(1/x) - 1 Convergent?
The area under curve f(x) = e^(1/x) - 1 gets smaller and smaller as x approaches infinity. Since the similar integral of e^(-x) from x = a to x = ∞ converges, it is natural to wonder if the integral of e^(1/x) - 1 from x = a to x = ∞ also converges. It turns out that the answer is no.
Since e^(1/x) - 1 is always greater than 1/x for values of x larger than 1, and the infinite integral of 1/x is divergent, then e^(1/x) - 1 is also divergent.