# about integration help 1,change of variable

$$\int_0^1 \frac {x - 1} {\ln x} \, dx =\, ?$$

I have not any idea how start it. Should I use change of variable? Or partial integration? $\frac 1 {\ln x}=u \rightarrow (x-1)dx=dv$ is it true? I cannot find conclusion.

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I think this problem can't be solved with the techniques you learn in high school calculus. –  DanielV Dec 1 '13 at 19:47
equivalent to $$\int_{-\infty}^{0} \frac{e^t-1}{te^t}dt$$ –  Suraj M S Dec 1 '13 at 19:57
Generally speaking, $$\int_0^1\frac{x^n-1}{\ln x}dx=\ln(n+1)$$ –  Lucian Dec 1 '13 at 20:40

Okay so I figured it out by digging in my old books. So there is one infamous represention of the natural logarithm as $$\ln(x) = -\lim_{n->0} \int_n^{\infty}\frac{e^{-xt}-e^{-t}}{t}dt$$ In your case if we let $u=-\ln(x)$ the integral becomes $$-\int_0^{\infty}\frac{e^{-2u}-e^{-u}}{u}du = \ln(2)$$ Equality holds as improper integral evaluation is defined as the limit of the proper.

To proove the first statement there are 2 major points:

1) $\ln(1) = 0$ and for $x=1$ the limit is $$-\lim_{n->0} \int_n^{\infty}\frac{e^{-xt}-e^{-t}}{t}dt=-\lim_{n->0} \int_n^{\infty}\frac{e^{-t}-e^{-t}}{t}dt = \lim_{n->0} \int_n^{\infty}0dt = 0$$

2) We proove that their derivatives with respect to $x$ are euqal: $$\frac{d\ln(x)}{dx} = \frac{d\left[-\lim_{n->0} \int_n^{\infty}\frac{e^{-xt}-e^{-t}}{t}dt\right]}{dx}$$

3)Since both of these are true $\forall x > 0$ by the fundamental theorem of calculus it follows that their equal up to a constant. Since we also proved and 1) it follows that they are exactly euqal. Write me down if you need a proof of point 2).

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Actually, the natural change of variable yields the result... Using $x=\mathrm e^{-t}$, one sees that the integral to be computed is $F(2)$, where, for every $a\gt1$, $$F(a)=\int_0^\infty\frac{\mathrm e^{-t}-\mathrm e^{-at}}t\,\mathrm dt.$$ The function $F$ is smooth on $(1,+\infty)$ and, for every $a\gt1$, $$F'(a)=\int_0^\infty\frac{\partial}{\partial a}\left(\frac{\mathrm e^{-t}-\mathrm e^{-at}}t\right)\,\mathrm dt=\int_0^\infty\mathrm e^{-at}\,\mathrm dt=\frac1a.$$ Since $\lim\limits_{a\to1}F(a)=0$, $F(a)=\log a$ for every $a\gt1$. In particular, $$\int_0^1 \frac {x - 1} {\ln x} \,\mathrm dx =\log2.$$

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The text passage contains a mathematical question in the form of an integral and provides a detailed solution to the problem, including multiple methods and explanations.
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First, describe the content structure you found in 1-2 sentences.

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# about integration help 1,change of variable

$$\int_0^1 \frac {x - 1} {\ln x} \, dx =\, ?$$

I have not any idea how start it. Should I use change of variable? Or partial integration? $\frac 1 {\ln x}=u \rightarrow (x-1)dx=dv$ is it true? I cannot find conclusion.

-
I think this problem can't be solved with the techniques you learn in high school calculus. –  DanielV Dec 1 '13 at 19:47
equivalent to $$\int_{-\infty}^{0} \frac{e^t-1}{te^t}dt$$ –  Suraj M S Dec 1 '13 at 19:57
Generally speaking, $$\int_0^1\frac{x^n-1}{\ln x}dx=\ln(n+1)$$ –  Lucian Dec 1 '13 at 20:40

Okay so I figured it out by digging in my old books. So there is one infamous represention of the natural logarithm as $$\ln(x) = -\lim_{n->0} \int_n^{\infty}\frac{e^{-xt}-e^{-t}}{t}dt$$ In your case if we let $u=-\ln(x)$ the integral becomes $$-\int_0^{\infty}\frac{e^{-2u}-e^{-u}}{u}du = \ln(2)$$ Equality holds as improper integral evaluation is defined as the limit of the proper.

To proove the first statement there are 2 major points:

1) $\ln(1) = 0$ and for $x=1$ the limit is $$-\lim_{n->0} \int_n^{\infty}\frac{e^{-xt}-e^{-t}}{t}dt=-\lim_{n->0} \int_n^{\infty}\frac{e^{-t}-e^{-t}}{t}dt = \lim_{n->0} \int_n^{\infty}0dt = 0$$

2) We proove that their derivatives with respect to $x$ are euqal: $$\frac{d\ln(x)}{dx} = \frac{d\left[-\lim_{n->0} \int_n^{\infty}\frac{e^{-xt}-e^{-t}}{t}dt\right]}{dx}$$

3)Since both of these are true $\forall x > 0$ by the fundamental theorem of calculus it follows that their equal up to a constant. Since we also proved and 1) it follows that they are exactly euqal. Write me down if you need a proof of point 2).

-

Actually, the natural change of variable yields the result... Using $x=\mathrm e^{-t}$, one sees that the integral to be computed is $F(2)$, where, for every $a\gt1$, $$F(a)=\int_0^\infty\frac{\mathrm e^{-t}-\mathrm e^{-at}}t\,\mathrm dt.$$ The function $F$ is smooth on $(1,+\infty)$ and, for every $a\gt1$, $$F'(a)=\int_0^\infty\frac{\partial}{\partial a}\left(\frac{\mathrm e^{-t}-\mathrm e^{-at}}t\right)\,\mathrm dt=\int_0^\infty\mathrm e^{-at}\,\mathrm dt=\frac1a.$$ Since $\lim\limits_{a\to1}F(a)=0$, $F(a)=\log a$ for every $a\gt1$. In particular, $$\int_0^1 \frac {x - 1} {\ln x} \,\mathrm dx =\log2.$$

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user
The text passage contains a mathematical question in the form of an integral and provides a detailed solution to the problem, including multiple methods and explanations.

TYPE: SINGLE_QUESTION_WITH_ANSWER