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Differentiating an accumulation function to recover ln(x)

Recall the natural logarithm can be defined as an accumulation function: F(x)=∫1x1t dtF(x) = \displaystyle\int_1^x \frac{1}{t}\,dt, for x>0x > 0.

Let F(x)=∫1x1t dt,x>0.F(x) = \int_1^x \frac{1}{t}\,dt, \quad x > 0. Use the Fundamental Theorem of Calculus, Part 1, to find F′(x)F'(x), and state which familiar function this shows F(x)F(x) must be (up to a constant).

Hint

FTC Part 1 says that if F(x)=∫axf(t) dtF(x) = \int_a^x f(t)\,dt with ff continuous, then F′(x)=f(x)F'(x) = f(x). Here the integrand is f(t)=1/tf(t) = 1/t and the upper limit is just xx itself, so no chain rule is needed.

Step-by-step solution

By FTC Part 1, since f(t)=1/tf(t) = 1/t is continuous on (0,∞)(0,\infty) and the upper limit of integration is simply xx, we directly get F′(x)=f(x)=1xF'(x) = f(x) = \dfrac{1}{x}.

Since ddx[ln⁡x]=1x\dfrac{d}{dx}[\ln x] = \dfrac{1}{x} as well, F(x)F(x) and ln⁡x\ln x have the same derivative on (0,∞)(0,\infty), so they differ by at most a constant. Evaluating at x=1x=1 gives F(1)=∫111t dt=0=ln⁡1F(1) = \int_1^1 \frac{1}{t}\,dt = 0 = \ln 1, so the constant is 00 and F(x)=ln⁡xF(x) = \ln x.

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