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In thermodynamics, I was trying to solve the following integral.

$$\int_{T_{1}}^{T_{2}} d \ln K=-\frac{\Delta H}{R} \int_{T_{1}}^{T_{2}} d\left(\frac{1}{T}\right)$$

$$\ln K\left(T_{2}\right)-\ln K\left(T_{1}\right)=-\frac{\Delta H}{R}\left(\frac{1}{T_{2}}-\frac{1}{T_{1}}\right)$$

$$\int_{T_{1}}^{T_{2}} d\left(\frac{1}{T}\right)$$

When I was studying this at school, normally we would have an indefinite integral such as

$$\int x dx$$ where we have a variable and we are integrating it with respect to a small change to (most of the time) that variable itself, i.e. integrate $x$ with respect to $dx$.

But in this $\int_{T_{1}}^{T_{2}} d\left(\frac{1}{T}\right)$ I was left rather stuck.

Are we integrating $1$ with respect to $d\left(\frac{1}{T}\right)$ or is this an incorrect definition of the statement?

I'm just a bit confused by what this means in Layman terms.

NOTE This is for an introductory chemistry student as well, apologies if the concept seems trivial to others.

vik1245
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  • Note that $x$ is just some symbol, as is $1/T$... although you may choose to, and people often do, ascribe more meaning to the second thing, in this case, pretend you have no idea about the symbols. Just integrate with respect to what is there. – The Count Jun 24 '19 at 00:19

3 Answers3

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The notation may become more intuitive if we look at the Riemann-sum's side. Indeed, much like the ordinary Riemann integral is simply

$$ \int_{a}^{b} f(x) \, \mathrm{d}x \approx \sum_{i=1}^{n} f(x_i) (x_i - x_{i-1}) $$

where the error becomes smaller as the finer partition $\{a = x_0 < x_1 < \cdots < x_n = b\}$ is considered, the Riemann-Stieltjes integral is

$$ \int_{a}^{b} f(x) \, \mathrm{d}g(x) \approx \sum_{i=1}^{n} f(x_i) (g(x_i) - g(x_{i-1})). $$

Here are particular cases:

  1. If $f$ is constant with the value $c$, then all the intermediate terms cancel out, yielding

    $$ \int_{a}^{b} c \, \mathrm{d}g(x) \approx \sum_{i=1}^{n} c (g(x_i) - g(x_{i-1})) = c(g(b) - g(a)). $$

  2. If $g$ is nice (continuously differentiable, for example), then $g(x_i) - g(x_{i-1}) \approx g'(x_i)(x_i - x_{i-1})$ by linear approximation, and so,

    $$ \int_{a}^{b} f(x) \, \mathrm{d}g(x) \approx \sum_{i=1}^{n} f(x_i) g'(x_i) (x_i - x_{i-1}) \approx \int_{a}^{b} f(x)g'(x) \, \mathrm{d}x. $$

Of course, the definition of Riemann-Stieltjes integral allows a more general class of increments $\mathrm{d}g(x)$ other than those proportional to $\mathrm{d}x$, which is both theoretically and computationally useful.

Community
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Sangchul Lee
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  • This answer is too excessive. The chief asked for Layman terms. – NestorV S Jun 24 '19 at 00:31
  • @NestorVS It is a correct answer, and provides the underlying meaning to the notation. Not all answers need to be tailored to the person who asked the question. Otherwise, there is no purpose in a community for sharing questions and answers, and no allowance for motivating mathematical growth. – heropup Jun 24 '19 at 09:30
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You're integrating 1 with respect to $\frac1T.$
To make it more simple to understand, let $u=\frac1T.$ Then $$\int_{T_1}^{T_2}d(\frac1T)=\int_{\frac{1}{T_1}}^{\frac{1}{T_2}}du=u\Big|_{\frac{1}{T_1}}^{\frac{1}{T_2}}=\frac{1}{T_2}-\frac{1}{T_1}$$

NestorV S
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note that: $$\frac{d}{dT}\left[\frac 1T\right]=-\frac{1}{T^2}$$ now if we rearrange we get: $$d\left[\frac 1T\right]=-\frac{dT}{T^2}$$ applying this to your integral we get: $$\int_{T_1}^{T_2}d\left[\frac 1T\right]=\int_{T_1}^{T_2}-\frac{dT}{T^2}=\left[\frac 1T\right]_{T_1}^{T_2}=\frac 1{T_2}-\frac{1}{T_1}$$

Henry Lee
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