Integral involving exponential

[stevendaryl]

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:

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

 

[anuttarasammyak]

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

 

[stevendaryl]

$$\frac{1}{1+\lambda x}=\sum_{j=0}^\infty (-\lambda x )^j$$
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 $\lambda$.

 

[fresh_42]

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:
https://www.physicsforums.com/threads/micromass-big-integral-challenge.867904/page-2#post-5451293

 

[Office_Shredder]

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

 

[DrClaude]

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]

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?

 

[DrClaude]

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.

 

[dextercioby]

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 $\text{Ei}\left(-\frac{1}{\lambda}\right)$

 

[romsofia]

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!