Find the indefinite integral of e^x/x with respect to x.

Step-by-step solution

1. 1

   Try the usual techniques and watch them fail. Substitution needs the derivative of the inner function to appear as a factor, and neither $u=x$ nor $u=e^x$ produces the $\tfrac1x$. Integration by parts only shuffles the difficulty: taking $u=e^x$, $dv=\tfrac{dx}{x}$ gives $e^x\ln x-\int e^x\ln x\,dx$, which is worse. This is a signal, not bad luck.

2. 2

   State the actual result. By Liouville's theorem on elementary integration, $\dfrac{e^x}{x}$ has no elementary antiderivative — no finite combination of polynomials, exponentials, logarithms, trigonometric functions and radicals differentiates to it. The antiderivative is instead named:

   $\int\frac{e^x}{x}\,dx=\operatorname{Ei}(x)+C$

3. 3

   Define the exponential integral. One standard definition is

   $\operatorname{Ei}(x)=\int_{-\infty}^{x}\frac{e^t}{t}\,dt$

   taken as a Cauchy principal value because the integrand blows up at $t=0$. By the Fundamental Theorem of Calculus, differentiating this definition returns the integrand:

   $\frac{d}{dx}\operatorname{Ei}(x)=\frac{e^x}{x}$

   which is exactly the property required of an antiderivative.

4. 4

   Note that a series is available if a formula is wanted. Expanding $e^x=\sum_{n\ge0}\tfrac{x^n}{n!}$ and dividing by $x$:

   $\frac{e^x}{x}=\frac1x+\sum_{n\ge1}\frac{x^{n-1}}{n!}$

   Integrating term by term gives

   $\int\frac{e^x}{x}\,dx=\ln|x|+\sum_{n\ge1}\frac{x^{n}}{n\cdot n!}+C$

   which equals $\operatorname{Ei}(x)-\gamma$ for $x>0$, with $\gamma\approx0.5772$ the Euler-Mascheroni constant. This is the practical way to compute values.

5. 5

   Contrast with the integrals that do work. $\int e^x\,dx=e^x+C$ and $\int\tfrac1x\,dx=\ln|x|+C$ are both elementary; it is only their quotient that escapes. Similarly $\int e^{-x^2}dx$ and $\int\tfrac{\sin x}{x}dx$ are non-elementary, while $\int xe^{x}dx$ is fine — there is no simple rule of thumb, which is why the theorem matters.

6. 6

   Sanity-check the series numerically. At $x=1$ the truncated series $\ln1+\sum_{n=1}^{20}\tfrac{1}{n\cdot n!}=1.31790$, and $\operatorname{Ei}(1)-\gamma=1.89512-0.57722=1.31790$ ✓ — the two agree, confirming the term-by-term integration.