# Thread: find asymptotic expansion of this integral

1. ## find asymptotic expansion of this integral

$f(x) = \int_{0}^{\infty}{\frac{e^{-t}}{(1+xt)^{2}} dt}$

as

$x\rightarrow 0^{+}$

2. Proceeding in 'trivial' way, if ...

$f(x)= \int_{0}^{\infty} \frac{e^{-t}\cdot dt}{(1+xt)^{2}}$ (1)

... You can compute...

$f(0) = \int_{0}^{\infty} e^{-t}\cdot dt =1$ (2)

What you are searching however is...

$\lim_{x \rightarrow 0+} f(x)$

... and this limit is 1 only if $f(x)$ is continuos in $x=0$. A way to meet that is to find an explicit expression for $f(x)$, what I will try to do now... lets hope with success! ...

We start by defining...

$\varphi(t)= \frac {1}{\frac{1}{x} + t}$ (3)

... and then we compute [using a textbook for saving time...] the LT of $\varphi(*)$ ...

$\int_{0}^{\infty} \varphi(t) \cdot e^{-st}\cdot dt = \mathcal {L} \{\varphi(t)\} = \mathcal {L} \{\frac{1}{\frac{1}{x}+t}\} = e^{\frac{s}{x}}\cdot Ei(\frac{s}{x})$ (4)

... where $Ei(*)$ is the socalled Integralexponential Function defined as...

$Ei(t)= \int_{t}^{\infty} \frac{e^{-u}\cdot du}{u}$ (5)

The next step is the computation of the LT of $\varphi^{'}(t)$ using the formula...

$\mathcal {L} \{\varphi^{'} (t)\} = s\cdot \mathcal {L} \{\varphi(t)\} - \varphi(0)$ (6)

Because is...

$\varphi^{'}(t)= - \frac{1}{(\frac{1}{x}+t)^{2}}$ (7)

... we obtain...

$- \int_{0}^{\infty} \frac{e^{-st}\cdot dt}{(\frac{1}{x} + t)^{2}}= s\cdot e^{\frac{s}{x}}\cdot Ei(\frac{s}{x}) - x$ (8)

Combining (1) and (8) computed for $s=1$ we finally obtain...

$f(x)= \int_{0}^{\infty} \frac{e^{-t}\cdot dt}{(1+xt)^{2}} = \frac{x - e^{\frac{1}{x}}\cdot Ei(\frac{1}{x})}{x^{2}}$ (9)

The 'experts' of MHF are kindly pleased to control the computation, because the 'old wolf' has lost most of his teeth ... if computation is correct the next step is computing the limit for $x \rightarrow 0+$ of (9)...

Kind regards

$\chi$ $\sigma$

P.S. : I had to drastically modify the post in order to correct many mistakes... very sorry!...

3. Originally Posted by pengchao1024
$f(x) = \int_{0}^{\infty}{\frac{e^{-t}}{(1+xt)^{2}} dt}$

as

$x\rightarrow 0^{+}$

We can make a substitution:

$1+xt=u \Longrightarrow dt=\frac{du}{x}\,,\;t=\frac{u-1}{x}\,,\;1\leq u\leq\infty$ , and we get:

$\int\limits_0^\infty\frac{e^{-t}}{(1+xt)^2}\;dt=\frac{e^{\frac{1}{x}}}{x}\int\li mits_1^\infty\frac{e^{-\frac{1}{x}u}}{u^2}\;du$

If I'm not wrong, the last integral is convergent by comparison with

$\int\limits_1^\infty\frac{du}{u^2}\;du=1$ , and since $\frac{e^\frac{1}{x}}{x}\xrightarrow [x\to 0] {}\infty$ then the whole thing asymptotically diverges to $+\infty$

...or I am wrong, of course.

Tonio

4. Even with some difficulties we are arrived to an explicit form of ...

$f(x)= \int_{0}^{\infty} \frac{e^{-t}\cdot dt}{(1+xt)^{2}} =\frac{x-e^{\frac{1}{x}}\cdot Ei(\frac{1}{x})}{x^{2}}$ (1)

The problem is now to find the limit of $f(x)$ when $x$ tends to $0+$. At this scope we use the change of variables $\tau=\frac{1}{x}$ so that is...

$\lim_{x \rightarrow 0+} f(x)= \lim_{\tau \rightarrow \infty} f(\tau)$

... where...

$f(\tau)= \tau-\tau^{2}\cdot e^{\tau}\cdot Ei(\tau)$ (2)

If $F(s) = \mathcal{L} \{f(\tau)\}$ , then for the 'Final value theorem' is...

$\lim_{\tau \rightarrow \infty} f(\tau) = \lim_{s \rightarrow 0+} s\cdot F(s)$ (3)

We now proceed with the following successive steps...

$\mathcal {L} \{Ei(\tau)\}= \frac{\ln (1+s)}{s}$ (4)

$\mathcal {L} \{e^{\tau}\cdot Ei(\tau)\} = \frac{\ln s}{s-1}$ (5)

$\mathcal {L} \{\tau^{2}\cdot e^{\tau}\cdot Ei(\tau)\} = \frac{d^{2}}{ds^{2}} \frac{\ln s}{s-1}= \frac{1}{s^{2}} - \frac{1}{(s-1)^{2}} - \frac{1}{s\cdot (s-1)^{4}} + 2\cdot \frac{\ln s}{(s-1)^{3}}$ (6)

$\mathcal {L} \{\tau -\tau^{2}\cdot e^{\tau}\cdot Ei(\tau)\} = F(s) = \frac{1}{(s-1)^{2}} + \frac{1}{s\cdot (s-1)^{4}} - 2\cdot \frac{\ln s}{(s-1)^{3}}$ (7)

The final conclusion is...

$\lim_{\tau \rightarrow \infty} f(\tau) = \lim_{s \rightarrow 0+} s\cdot F(s) = 1$ (8)

... so that is...

$\lim_{x \rightarrow 0+} f(x) = f(0) = 1$ (9)

Kind regards

$\chi$ $\sigma$

5. Originally Posted by tonio
$\int\limits_0^\infty\frac{e^{-t}}{(1+_xt)^2}\;dt=\frac{e^{\frac{1}{x}}}{x}\int\l imits_1^\infty\frac{e^{-\frac{1}{x}u}}{u^2}\;du$

If I'm not wrong, the last integral is convergent by comparison with

$\int\limits_1^\infty\frac{du}{u^2}\;du=1$ , and since $\frac{e^\frac{1}{x}}{x}\xrightarrow [x\to \infty] {}\infty$ then the whole thing asymptotically diverges to $+\infty$
The problem is that the last integral converges to 0 when $x\to 0^+$, hence an undeterminate form $\infty\times 0$, you can't conclude.

Originally Posted by chisigma
The final conclusion is...

$\lim_{\tau \rightarrow \infty} f(\tau) = \lim_{s \rightarrow 0+} s\cdot F(s) = 1$ (8)

... so that is...

$\lim_{x \rightarrow 0+} f(x) = f(0) = 1$ (9)

Kind regards

$\chi$ $\sigma$
If one only needs $\lim_{x\to 0^+} f(x)$, then there is (much) simpler:

-On one hand, $f(x)\leq \int_0^\infty e^{-t}dt=1$.

-On the other hand, for any $A>0$, $f(x)\geq\int_0^A\frac{e^{-t}}{(1+xt)^2}dt \geq \frac{1}{(1+xA)^2}\int_0^A e^{-t}dt$ $=\frac{1-e^{-A}}{(1+Ax)^2}$. Choose $A(x)=\frac{1}{\sqrt{x}}$, so that $A(x)\to_{x\to 0^+} +\infty$ and $A(x)x=\sqrt{x}\to_{x\to 0^+} 0$, hence the previous lower bound tends to 1 as $x\to 0^+$.

--
Of course, the bounded convergence theorem gives a far less elementary, yet simpler proof: for all $x,t>0$, $0\leq \frac{e^{t}}{(1+xt)^2}\leq \varphi(t)=e^{-t}$ and $\int_0^\infty e^{-t}dt<\infty$, and the function in the integral tends to $e^{-t}$ when $x\to 0^+$, hence $f(x)\to_{x\to 0^+} \int_0^\infty e^{-t}dt =1$.

--
The original post asked for an asymptotic expansion, i.e. something like :

$f(x)=1-2x+6x^2-24 x^3+o_{x\to 0^+}(x^3)$

and more generally a pattern for the expansion.

I couldn't find a direct proof. Maybe ChiSigma's explicit formula could help.

What I did however is use the Taylor-Young expansion: $f(x)=f(0)+f'(0)x+\frac{f''(0)}{2}x^2+\cdots+\frac{ f^{(k)}(0)}{k!}x^k +o(x^k)$ if $f$ is sufficiently smooth near $x$.

Using the (non-elementary) theorem of derivation under the integral sign, we can get the $k$-th derivative of $f$ at $x>0$:

$f^{(k)}(x)=(-1)^k (k+1)! \int_0^\infty \frac{t^k}{(1+tx)^{2+k}}e^{-t} dt$

Hence (using the bounded convergence theorem like above)

$\lim_{x\to 0^+} f^{(k)}(x) = (-1)^k (k+1)! \int_0^\infty t^k e^{-t} dt = (-1)^k (k+1)! k!$.

We deduce (this is a theorem) that $f$ is indefinitely derivable at 0 (to the right) and

$f^{(k)}(0) = (-1)^k (k+1)! k!$.

Using Taylor-Young expansion at $0$ (to the right), we get

$f(x)=\sum_{k=0}^n (-1)^k (k+1)! x^k + o_{x\to 0^+}(x^n)$

for all $n\in\mathbb{N}$. (Of course, this is not a series expansion since it would be divergent)

(Mistakes are possible, I couldn't check this expansion)

6. A nice general purpose formula I found some years ago for the computation of integrals of the type...

$\int_{0}^{\infty} f(t)\cdot e^{-t}\cdot dt$

... is illustated. A well known formula that holds for any real or complex $t$ is...

$e^{-t} = \lim_{k \rightarrow \infty} (1-\frac{t}{k})^{k}$ (1)

On the basis of (1) it is 'spontaneous' to write...

$\int_{0}^{\infty} f(t)\cdot e^{-t}\cdot dt = \lim_{k \rightarrow \infty} \int_{0}^{k} f(t)\cdot (1-\frac{t}{k})^{k}\cdot dt$ (2)

... that with the change of variables $u=\frac{t}{k}$ becomes...

$\int_{0}^{\infty} f(t)\cdot e^{-t}\cdot dt = \lim_{k \rightarrow \infty} k \cdot \int_{0}^{1} f(ku)\cdot (1-u)^{k}\cdot du$ (3)

Now I'll try to use (3) for solving the problem proposed in this thread and I suggest You to do the same. I don't promise that my 'magic formula' will work in this case ... at the time it worked very well for the computation of the integral proposed in...

Kind regards

$\chi$ $\sigma$

7. In order to simplify the computation we trasform the integral as...

$f(x)= \int_{0}^{\infty} \frac{e^{-t}\cdot dt}{(1+tx)^{2}}= \frac{1}{x^{2}} \int_{0}^{\infty} \frac{e^{-t}\cdot dt}{(t+\frac{1}{x})^{2}}$ (1)

Now we apply the 'magic formula'...

$\int_{0}^{\infty} f(t)\cdot e^{-t}\cdot dt = \lim_{k \rightarrow \infty} k \cdot \int_{0}^{1} f(ku)\cdot (1-u)^{k}\cdot du$ (2)

... to the integral in (1) and obtain...

$\int_{0}^{\infty} \frac{e^{-t}\cdot dt}{(t+\frac{1}{x})^{2}}= \lim_{k \rightarrow \infty} \frac{1}{k} \cdot \int_{0}^{1} \frac{(1-u)^{k}}{(u+\frac{1}{kx})^{2}}\cdot du$ (3)

Now we proceed to integrate...

$\int_{0}^{1} \frac{(1-u)^{k}}{(u+\frac{1}{kx})^{2}}\cdot du = kx - \int_{0}^{1}\frac{(1-u)^{k-1}}{u+\frac{1}{kx}}\cdot du=$

$= kx -\ln (kx) - (k-1) \int_{0}^{1} (1-u)^{k-2} \cdot \ln (u+\frac{1}{kx})\cdot du=$

$=kx -\ln (kx) - \frac{k-1}{kx}\cdot \ln (kx) + (k-1)\cdot (k-2) \cdot \int_{0}^{1} u\cdot (1-u)^{k-3}\cdot du +$

$- (k-1)\cdot (k-2)\cdot \int_{0}^{1} (u+\frac{1}{kx})\cdot \ln (u+\frac{1}{kx})\cdot (1-u)^{k-3}\cdot du$ (4)

At this point a little stop is useful to do some considerations...

a) because is...

$\lim_{k \rightarrow \infty} \frac{\ln (kx)}{k} = \lim_{k \rightarrow \infty} \frac{(k-1)\cdot \ln (kx)}{k^{2}\cdot x} = 0$ (5)

... the terms containing $\ln (kx)$ can be deleted from (4)...

b) in order to proceed it is very useful the 'little nice formula'...

$\int_{0}^{1} u^{n}\cdot (1-u)^{m}\cdot du = \frac {n!}{(m+1)\cdot (m+2)\dots (m+n+1)} = \frac{1}{\binom{m+n+1}{n}}$ (6)

Now if we set in (6) $n=1$ and $m= k-3$ we have...

$(k-1)\cdot (k-2) \cdot \int_{0}^{1} u\cdot (1-u)^{k-3}\cdot du = 1$ (7)

... and because is...

$\lim_{k \rightarrow \infty} \frac{1}{k} =0$ (8)

... also this term can be deleted from (4). In conclusion we can 'simplify' (4) writing...

$\int_{0}^{\infty} \frac{e^{-t}\cdot dt}{(t+\frac{1}{x})^{2}}= x -\lim_{k \rightarrow \infty} \frac{(k-1)\cdot (k-2)}{k} \int_{0}^{1} (u+\frac{1}{kx})\cdot \ln (u+\frac{1}{kx})\cdot (1-u)^{k-3}\cdot du$ (9)

Before proceeding, the 'experts' of MHF are kindly requested to verify the computation I have done till now... thanks a lot! ...

Kind regards

$\chi$ $\sigma$

8. Originally Posted by chisigma
In order to simplify the computation we trasform the integral as...

$f(x)= \int_{0}^{\infty} \frac{e^{-t}\cdot dt}{(1+tx)^{2}}= \frac{1}{x^{2}} \int_{0}^{\infty} \frac{e^{-t}\cdot dt}{(t+\frac{1}{x})^{2}}$ (1)

Now we apply the 'magic formula'...
(...)
$\int_{0}^{\infty} \frac{e^{-t}\cdot dt}{(t+\frac{1}{x})^{2}}= x -\lim_{k \rightarrow \infty} \frac{(k-1)\cdot (k-2)}{k} \int_{0}^{1} (u+\frac{1}{kx})\cdot \ln (u+\frac{1}{kx})\cdot (1-u)^{k-3}\cdot du$ (9)
I'd say I'm rather skeptical about your chances to conclude. It would have been encouraging if you had obtained something like $f(x)=1-2x+\lim_k\int_0^1\phi(k,x)dx$ since it could suggest (refering to the expansion I obtained) that the integral is negligible with respect to $x$ and that you'll get the whole asymptotic expansion by a - possibly tiresome - induction. However you get something like $f(x)=\frac{1}{x}+\lim_k \int_0^1\phi(k,x)dx$, hence now you'll have to prove that the second term contains a term $-\frac{1}{x}$ to cancel the first one; thus it doesn't look like this new formula is any "better" than that you started with. Plus there may be a difficulty when you'll have to exchange limits $x\to0^+$ and $k\to\infty$.

That's a drawback with magic formulas: either they really act like magic, or they obfuscate things since we have to apply them "at random" in hope they will work. The down-to-earth arguments of my post (i.e. derivation and Taylor-Young expansion), while not especially beautiful, have the advantage to work for sure. And to be short and simple.

9. ## Non c'è due senza tre...

An old italian proverb say : '... non c'è due senza tre!...', that's in english idiom is [approximatively...]: '... never two without a third!...'. In this thread I've showed to You two 'nice little formulas'...

$\int_{0}^{\infty} f(t)\cdot e^{-t}\cdot dt = \lim_{k \rightarrow \infty} k \cdot \int_{0}^{1} f(ku)\cdot (1-u)^{k}\cdot du$ (1)

$\int_{0}^{1} u^{n}\cdot (1-u)^{m}\cdot du = \frac{n!}{(m+1)\cdot (m+2)\dots (m+n+1)} = \frac{1}{\binom{m+n+1}{n}}$ (2)

Since tomorrow is Halloween's holiday , I'll donate to You a third 'nice little formula' . Let suppose we have to compute the indefinite integral...

$\int (t+a)^{m} \cdot \ln^{n} (t+a)\cdot dt$ (3)

Ai first it seems a little hard... never mind, we proceed 'step by step' integrating first (3) by parts...

$\int (t+a)^{m} \cdot \ln^{n} (t+a)\cdot dt = \frac{(t+a)^{m+1}}{m+1} \cdot \ln^{n} t - \frac{n}{m+1}\cdot \int (t+a)^{m}\cdot \ln^{n-1} (t+a)\cdot dt$ (4)

The result is that now we have an integral similar to the first... but simpler, because the exponent of the logarithm is $n-1$ instead of $n$!... very good !... we can proceed till to have the exponent of the logarithm equal to 0 and obtain...

$\int (t+a)^{m} \cdot \ln^{n} (t+a)\cdot dt = (t+a)^{m+1} \cdot \sum_{i=0}^{n} \frac{(-1)^{i}}{(m+1)^{i+1}}\cdot \frac{n!}{(n-i)!}\cdot \ln^{n-i} (t+a) + c$ (5)

Very nice indeed! ... and I hope it will be also very useful! ...

Have You all a nice Halloween!...

Kind regards

$\chi$ $\sigma$

10. Today is Halloween holiday and I will offer You a nice and useful present: the generalization of one of the ‘nice little formulas’ we have seen in previous posts

Let suppose we have to compute the integral…

$\int_{0}^{1} (u+a)^{n}\cdot (1-u)^{m}\cdot du$ (1)

A good idea is, of course, the integration by parts and after one step we obtain...

$\int_{0}^{1} (u+a)^{n}\cdot (1-u)^{m}\cdot du = \frac{1}{m+1}\cdot a^{n} + \frac{n}{m+1}\cdot \int_{0}^{1} (u+a)^{n-1}\cdot (1-u)^{m+1}\cdot du$ (2)

The idea seems to work well, so that we try a second step...

$\int_{0}^{1} (u+a)^{n}\cdot (1-u)^{m}\cdot du = \frac{1}{m+1}\cdot a^{n} + \frac{n}{(m+1)\cdot (m+2)} \cdot a^{n-1} +$

$+ \frac{n\cdot (n-1)}{(m+1)\cdot (m+2)}\cdot \int_{0}^{1} (u+a)^{n-2}\cdot (1-u)^{m+2}\cdot du$ (3)

At this point it is 'spontaneous' to proceed along this way till to obtain...

$\int_{0}^{1} (u+a)^{n}\cdot (1-u)^{m}\cdot du = \frac{a^{n}}{m+1} +\sum_{i=1}^{n} \frac{n\cdot (n-1)\dots (n-i+1)}{(m+1)\cdot (m+2)\dots (m+i+1)}\cdot a^{n-i}$ (4)

Les jeaux son fait!... rien ne va plus! ...

Kind regards

$\chi$ $\sigma$