# Integral involving exponential

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• stevendaryl
In summary, If ##\lambda > 0##, the integral converges, but it's not analytic in ##\lambda##. If you set ##t \equiv 1/\lambda + x##, you get $$F(\lambda) = \frac{e^{1/\lambda}}{\lambda} \int_{1/\lambda}^{\infty} \frac{e^{-t}}{t} dt = \frac{e^{1/\lambda}}{\lambda} \Gamma(0, 1/\lambda)$$

#### stevendaryl

Staff Emeritus
Just a quick question:

Does anybody know if there is a closed-form solution to this rather simple-looking definite integral?

##F(\lambda) = \int_0^{\infty} \dfrac{e^{-x}}{1 + \lambda x} dx##

If ##\lambda > 0##, it definitely converges. It has a limit of 1 as ##\lambda \rightarrow 0##. But it doesn't seem to be analytic in ##\lambda##, since if you try to do a power series in ##\lambda##, you get a nonconvergent sequence:

##F(\lambda) = \sum_{j=0}^\infty (-1)^j \lambda^j (j!)##

$$\frac{1}{1+\lambda x}=\sum_{j=0}^\infty (-\lambda x )^j$$
only if ##|\lambda x| < 1##?

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anuttarasammyak said:
$$\frac{1}{1+\lambda x}=\sum_{j=0}^\infty (-\lambda x )^j$$
only if ##|\lambda x| < 1## ?

Yes, the summation doesn't converge if ##|\lambda x| > 1##. It's even worse if you put the factorials in:

##\sum_{j=0}^\infty (-\lambda)^j j!##

doesn't converge for any nonzero value of ##\lambda##.

stevendaryl said:
Just a quick question:

Does anybody know if there is a closed-form solution to this rather simple-looking definite integral?

##F(\lambda) = \int_0^{\infty} \dfrac{e^{-x}}{1 + \lambda x} dx##

If ##\lambda > 0##, it definitely converges. It has a limit of 1 as ##\lambda \rightarrow 0##. But it doesn't seem to be analytic in ##\lambda##, since if you try to do a power series in ##\lambda##, you get a nonconvergent sequence:

##F(\lambda) = \sum_{j=0}^\infty (-1)^j \lambda^j (j!)##
It looks like the Gamma function. See 1.1 / 1.8 / 2.3 in https://de.wikibooks.org/wiki/Formelsammlung_Mathematik:_Bestimmte_Integrale:_Form_R(x,exp)
or the integration trick used here:

If ##\lambda <0## then the integral is not defined, so you generically should not expect to get a power series around 0 that works.

stevendaryl
Setting ##t \equiv 1/\lambda + x##, you get
$$F(\lambda) = \frac{e^{1/\lambda}}{\lambda} \int_{1/\lambda}^{\infty} \frac{e^{-t}}{t} dt = \frac{e^{1/\lambda}}{\lambda} \Gamma(0, 1/\lambda)$$
where ##\Gamma(s, x)## is the incomplete gamma function.

romsofia, dextercioby, Infrared and 3 others
DrClaude said:
Setting ##t \equiv 1/\lambda + x##, you get
$$F(\lambda) = \frac{e^{1/\lambda}}{\lambda} \int_{1/\lambda}^{\infty} \frac{e^{-t}}{t} dt = \frac{e^{1/\lambda}}{\lambda} \Gamma(0, 1/\lambda)$$
where ##\Gamma(s, x)## is the incomplete gamma function.
If you don't mind sharing, how did you notice this? Was it because you saw a similar integral, or was it a guess/intuition?

romsofia said:
If you don't mind sharing, how did you notice this? Was it because you saw a similar integral, or was it a guess/intuition?
I started from the solution Mathematica gave to the integral and then played around until I could understand how to get such a solution.

romsofia
romsofia said:
If you don't mind sharing, how did you notice this? Was it because you saw a similar integral, or was it a guess/intuition?

Using "special functions" is a way of cheating, since all the theory of integrals in terms of them is already embedded either in a book like Abramowitz-Stegun/Gradshteyn-Ryzhik, or in a computer software to which G-S has already been fed + it was taught various substitution techniques.

Therefore, when you see the integral and in the absence of a computer software you can feed the problem to,
you can try to devise certain substitutions to bring it to a form already present in G-S.

Here is how it's done.
https://en.wikipedia.org/wiki/Exponential_integral
In order to bring it to the https://en.wikipedia.org/wiki/Exponential_integral, you need to first substitute

$$1 + \lambda x = t$$

Then the integral is just a numerical factor times ##\text{Ei}\left(-\frac{1}{\lambda}\right)##

dextercioby said:
Using "special functions" is a way of cheating, since all the theory of integrals in terms of them is already embedded either in a book like Abramowitz-Stegun/Gradshteyn-Ryzhik, or in a computer software to which G-S has already been fed + it was taught various substitution techniques.

Therefore, when you see the integral and in the absence of a computer software you can feed the problem to,
you can try to devise certain substitutions to bring it to a form already present in G-S.

Here is how it's done.
https://en.wikipedia.org/wiki/Exponential_integral
In order to bring it to the https://en.wikipedia.org/wiki/Exponential_integral, you need to first substitute

$$1 + \lambda x = t$$

Then the integral is just a numerical factor times ##\text{Ei}\left(-\frac{1}{\lambda}\right)##
I need to get more comfortable using those books, I remember my physics teacher in high school showing me that one with the red cover. But thanks for the insight!