Introduction to the exponential integrals
The exponential‐type integrals have a long history. After the early developments of differential calculus, mathematicians tried to evaluate integrals containing simple elementary functions, especially integrals that often appeared during investigations of physical problems. Despite the relatively simple form of the integrands, some of these integrals could not be evaluated through known functions. Examples of integrals that could not be evaluated in known functions are:
L. Euler (1768) introduced the first integral shown in the preceding list. Later L. Mascheroni (1790, 1819) used it and introduced the second and third integrals, and A. M. Legendre (1811) introduced the last integral shown. T. Caluso (1805) used the first integral in an article and J. von Soldner (1809) introduced its notation through symbol li. F. W. Bessel (1812) used the second and third integrals. C. A. Bretschneider (1843) not only used the second and third integrals, but also introduced similar integrals for the hyperbolic functions:
O. Schlömilch (1846) and F. Arndt (1847) widely used such integrals containing the exponential and trigonometric functions. For the exponential, sine, and cosine integrals, J. W. L. Glaisher (1870) introduced the notations , , and . H. Amstein (1895) introduced the branch cut for the logarithmic integral with a complex argument. N. Nielsen (1904) used the notations and for corresponding integrals.
Different notations are used for the previous definite integrals by various authors when they are integrated from to or from to .
Definitions of exponential integrals
The exponential integral , exponential integral , logarithmic integral , sine integral , hyperbolic sine integral , cosine integral , and hyperbolic cosine integral are defined as the following definite integrals, including the Euler gamma constant :
The previous integrals are all interrelated and are called exponential integrals.
Instead of the above classical definitions through definite integrals, equivalent definitions through infinite series can be used, for example, the exponential integral can be defined by the following formula (see the following sections for the corresponding series for the other integrals):
A quick look at the exponential integrals
Here is a quick look at the graphics for the exponential integrals along the real axis.
Connections within the group of exponential integrals and with other function groups
Representations through more general functions
The exponential integrals , , , , , , and are the particular cases of the more general hypergeometric and Meijer G functions.
For example, they can be represented through hypergeometric functions or the Tricomi confluent hypergeometric function :
Representations of the exponential integrals and , the sine and cosine integrals and , and the hyperbolic sine and cosine integrals and through classical Meijer G functions are rather simple:
Here is the Euler gamma constant and the complicated‐looking expression containing the two logarithm simplifies piecewise:
But the last four formulas that contain the Meijer G function can be simplified further by changing the classical Meijer functions to the generalized one. These formulas do not include factors and terms :
The corresponding representations of the logarithmic integral through the classical Meijer G function is more complicated and includes composition of the G function and a logarithmic function:
All six exponential integrals of one variable are the particular cases of the incomplete gamma function:
Representations through related equivalent functions
The exponential integral can be represented through the incomplete gamma function or the regularized incomplete gamma function:
Relations to inverse functions
The exponential integral is connected with the inverse of the regularized incomplete gamma function by the following formula:
Representations through other exponential integrals
The exponential integrals , , , , , , and are interconnected through the following formulas:
The best-known properties and formulas for exponential integrals
Real values for real arguments
For real values of parameter and positive argument , the values of the exponential integral are real (or infinity). For real values of argument , the values of the exponential integral , the sine integral , and the hyperbolic sine integral are real. For real positive values of argument , the values of the logarithmic integral , the cosine integral , and the hyperbolic cosine integral are real.
Simple values at zero
The exponential integrals have rather simple values for argument :
Specific values for specialized parameter
If the parameter equals , the exponential integral can be expressed through an exponential function multiplied by a simple rational function. If the parameter equals , the exponential integral can be expressed through the exponential integral , and the exponential and logarithmic functions:
The previous formulas are the particular cases of the following general formula:
If the parameter equals , the exponential integral can be expressed through the probability integral , and the exponential and power functions, for example:
The previous formulas can be generalized by the following general representation of this class of particular cases:
The exponential integrals , , , , , , and are defined for all complex values of the parameter and the variable . The function is an analytical functions of and over the whole complex ‐ and ‐planes excluding the branch cut on the ‐plane. For fixed , the exponential integral is an entire function of . The sine integral and the hyperbolic sine integral are entire functions of .
Poles and essential singularities
For fixed , the function has an essential singularity at . At the same time, the point is a branch point for generic . For fixed , the function has only one singular point at . It is an essential singular point.
The exponential integral , the cosine integral , and the hyperbolic cosine integral have an essential singularity at .
The function does not have poles and essential singularities.
The sine integral and the hyperbolic sine integral have an essential singularity at .
Branch points and branch cuts
For fixed , the function does not have branch points and branch cuts.
For fixed , not being a nonpositive integer, the function has two branch points and , and branch cuts along the interval . At the same time, the point is an essential singularity for this function.
The exponential integral , the cosine integral, and the hyperbolic cosine integral have two branch points and .
The function has three branch points , , and .
The sine integral and hyperbolic sine integral do not have branch points or branch cuts.
For fixed , not being a nonpositive integer, the function is a single‐valued function on the ‐plane cut along the interval , where it is continuous from above:
The function is a single‐valued function on the ‐plane cut along the interval , where it has discontinuities from both sides:
The function is a single‐valued function on the ‐plane cut along the interval . It is continuous from above along the interval and it has discontinuities from both sides along the interval :
The cosine integral and hyperbolic cosine integral are single‐valued functions on the ‐plane cut along the interval
The standard method (typically found in first-year calculus textbooks) for determining the integrals
is to integrate by parts twice. If you haven’t seen the standard method, I’ll show you how to do the first one; the second one is similar. Later in the post I’ll show you the neat trick for determining both integrals at once.
It doesn’t much matter whether you let u represent the exponential function or the trigonometric function in the first integration, but you have to be consistent in the second integration. (That is, if you let u stand for the exponential function in the first integration, then also let u stand for the exponential function in the second integration. Alternatively, if you let u stand for the trigonometric function in the first integration, then also let u stand for the trigonometric function in the second integration.) Otherwise, after two integrations by parts you will end up with 0 = 0, which is true but not helpful.
So, let and , so that and . Then, calling the integral to be determined “I,” for reasons that will become clear shortly, we have
In the integral on the right side of the previous equation, integrate by parts again, letting and , so that and . The result is
We seem to be going around in circles, because the integral on the right side of the previous equation is the same as the one we started with. However, if we just replace it by its label I, the previous equation is seen to be an algebraic equation that we can solve for I. (This is the motivation for introducing the label I.) Doing this, we obtain
You can check the result by differentiating it to arrive at the original function. Some people enjoy this method as it seems as if we got something for nothing. We never really “finished” the integration by parts (after two iterations, we were still left with an integral), and yet the final result somehow popped out.
And of course, if you need to determine the other integral, then you have to go through the process once more, integrating by parts twice again. Try it for practice, if you wish; the result is
Now for the trick, which relies on you knowing about complex numbers, including Euler’s formula:
The strategy is to multiply the second of the original integrals by and then add it to the first of the original integrals. It turns out that combining them in this way results in an integral that is quite easy to determine; no integration by parts four times is needed. Then we just separate the final result into a sum of real and imaginary parts; the real part is the result for the first integral and the imaginary part is the result of the second integral.
Let’s see how it works: First multiply by and add it to and combine the integrals into one integral:
(using Euler’s formula)
Now combine the exponential functions on the right side of the previous equation and antidifferentiate:
The final stage is to express the right side of the previous equation as the sum of a real part and an imaginary part. Part of this process is to multiply the numerator and denominator of the fractional factor by the complex conjugate of the denominator; the other part is to separate the exponential functions on the right side of the equation:
Matching real and imaginary parts on both sides of the previous equation gives us the final results:
Conclusion: Does the trick save work? The integration step in the trick method is very easy, so we’re trading four integrations by parts plus some algebra for a simple integration and some algebra with complex numbers. It’s a trade I’d make any day. But of course, to each his own, so try both ways and decide for yourself which way you like better.
If the integrals are a bit more complex, then the savings in the trick method are even greater. For example, you might try using both methods to determine the integrals
Integration by parts is even more of a pain, but the trick method is hardly more difficult. If you do try them, you can check your final results against these:
I just did these two integrals with pencil and paper, and the trick method is much faster. Even if you only have to work out one of the integrals (the method is the same, you just ignore one of the final results), I think the trick method is still a time-saver and the probability of making an error is reduced, because you avoid the messy integrations by parts.
About Santo D'AgostinoI have taught mathematics and physics since the mid 1980s. I have also been a textbook writer/editor since then. Currently I am working independently on a number of writing and education projects while teaching physics at my local university. I love math and physics, and love teaching and writing about them. My blog also discusses education, science, environment, etc. https://qedinsight.wordpress.com Further resources, and online tutoring, can be found at my other site http://www.qedinfinity.com
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