Logarithmic and exponential integrals

In summary, the conversation discusses the definitions of the exponential and logarithmic integrals and the process of obtaining their power series expansions. It is mentioned that the exponential integral may only be meaningful in the sense of a Cauchy principal value and the Taylor series can be used to obtain the expansion for both positive and negative values of x.
  • #1
Millennial
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In this text, I will ask a question about the power series expansion of exponential and logarithmic integrals.

Now, to avoid confusion, I will first give the definitions of the two:
[tex]\mathrm{Ei}(x)=\int_{-\infty}^{x}\frac{e^t}{t}dt[/tex]
[tex]\mathrm{Li}(x)=\int_{0}^{x}\frac{dt}{\log(t)}[/tex]
where Ei denotes the exponential integral function and Li denotes the logarithmic integral function.

I think it is clear from now that these integrals are not expressable in elementary terms as indefinite ones. My question is about expressing these as power series.

Now, one can substitute [itex]t=e^u[/itex] in the logarithmic integral to get
[tex]\mathrm{Li}(x)=\int_{0}^{x}\frac{dt}{\log(t)}=\int_{-\infty}^{\log(x)}\frac{e^u}{u}du=\mathrm{Ei}(\log(x))[/tex]

so my question can be downgraded to only the power series expansion of the exponential integral rather than both. My question should be obvious by now: How do we obtain a power series expansion for the exponential integral, since term-by-term integration does not work after plugging in the Taylor series for the exponential function?
 
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  • #2
Millennial said:
[tex]\mathrm{Ei}(x)=\int_{-\infty}^{x}\frac{e^t}{t}dt[/tex]
For x > 0, I believe Ei is only meaningful in the sense of a Cauchy principal value. So maybe you can break it into
[tex]\mathrm{Ei}(x)=\int_{-\infty}^{-x}\frac{e^t}{t}dt + \int_{-x}^{x}\frac{e^t}{t}dt[/tex]
[tex]\int_{-x}^{x}\frac{e^t}{t}dt = \int_{0}^{x}\frac{e^t-e^{-t}}{t}dt[/tex]
Then the Taylor series should work for (et-e-t)/t.
The remaining portion is the same as the x < 0 case, which by swapping sign becomes:
[tex]\int_{x}^{\infty}\frac{e^{-t}}{t}dt = \int_{1}^{\infty}\frac{e^{-t}}{t}dt - \int_{1}^{x}\frac{e^{-t}}{t}dt[/tex]
Now the Taylor expansion works in the second integral. The first integral is some constant.
 

1. What is the difference between logarithmic and exponential integrals?

Logarithmic and exponential integrals are both types of mathematical functions that involve the use of logarithms and exponential functions. However, the main difference between the two is that logarithmic integrals involve the integration of logarithmic functions, while exponential integrals involve the integration of exponential functions.

2. What is the purpose of using logarithmic and exponential integrals?

Logarithmic and exponential integrals are useful in solving various mathematical problems, particularly in calculus and physics. They can also be used to model real-world phenomena, such as population growth and radioactive decay.

3. How do you solve logarithmic and exponential integrals?

To solve logarithmic and exponential integrals, you can use various techniques such as integration by parts, substitution, or trigonometric substitution. It is also helpful to have a good understanding of logarithmic and exponential rules and properties.

4. Are logarithmic and exponential integrals related in any way?

Yes, logarithmic and exponential integrals are related through the use of the inverse function property. This means that when the logarithmic and exponential functions are composed, they cancel each other out, making it easier to solve integrals involving both types of functions.

5. How are logarithmic and exponential integrals used in real life?

Logarithmic and exponential integrals have numerous applications in real life. For example, they are used in finance to calculate compound interest, in biology to model population growth, and in physics to describe exponential decay. They are also essential in the field of engineering and technology, particularly in signal processing and control systems.

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