Series expansion for an integral

Question

Working on a problem,

I tried to construct a series expansion for the following integral: $\displaystyle \ \int_{2}^{\sqrt{x}} \frac{1}{tlog(t)log(\frac{x}{t})}dt$;

I think that it would be: $\displaystyle \frac{log(log(x))}{log(x)}+ O\left(\frac{1}{log(x)}\right)$.

But I fails to get this estimate (for example using integration by parts)..how I can do this?

In general how I can find a series expansion for an integral of the form: $\displaystyle \ \int_{a}^{f(x)} g(x,t)dt$?

Answer 1

Too long for a comment.

Interesting enough, the antiderivative exists and it is even quite simple $$I=\int\frac{dt}{t \log (t) \log \left(\frac{x}{t}\right)}=\frac{\log (\log (t))-\log \left(\log \left(\frac{x}{t}\right)\right)}{\log (x)}$$ which makes $$J =\int_2^{\sqrt x}\frac{dt}{t \log (t) \log \left(\frac{x}{t}\right)}=\frac{\log \left(\frac{\log (x)}{\log (2)}-1\right)}{\log (x)}$$

Back to your initial question, you can use, around $t=\sqrt{x}$, the Taylor expansion of the integrand and get for it $$\frac{1}{\sqrt{x} \log ^2\left(\sqrt{x}\right)}-\frac{t-\sqrt{x}}{x \log ^2\left(\sqrt{x}\right)}+\frac{\left(t-\sqrt{x}\right)^2 \left(\log ^2\left(\sqrt{x}\right)+1\right)}{x^{3/2} \log ^4\left(\sqrt{x}\right)}+O\left(\left(t-\sqrt{x}\right)^3\right)$$ Integrating one term at the time is quite simple and, for the definite integral, you should arrive to something like

$$\frac{2 \left(\sqrt{x}-2\right) \left(8 \left(\sqrt{x}-2\right)^2+\left(11 x-14 \sqrt{x}+8\right) \log ^2(x)\right)}{3 x^{3/2} \log ^4(x)}$$

Let us try using $x=9$. The exact result is given by $$\frac{\log \left(\frac{\log \left(\frac{9}{2}\right)}{\log (2)}\right)}{\log (9)}\approx 0.352578$$ while the truncated Taylor series would lead to $$\frac{2 \left(8+65 \log ^2(9)\right)}{81 \log ^4(9)}\approx 0.340912$$

The table below shows the numerical value of the integral as a function of the number of terms $n$ used in the Taylor expansion $$\left( \begin{array}{cc} n & \text{value} \\ 1 & 0.3222082304 \\ 2 & 0.3409120137 \\ 3 & 0.3477066971 \\ 4 & 0.3505046461 \\ 5 & 0.3516727070 \\ 6 & 0.3521766148 \\ 7 & 0.3523976036 \\ 8 & 0.3524960302 \\ 9 & 0.3525403739 \\ 10 & 0.3525605467 \\ 11 & 0.3525697972 \\ 12 & 0.3525740681 \\ 13 & 0.3525760515 \\ 14 & 0.3525769772 \\ 15 & 0.3525774111 \\ 16 & 0.3525776154 \\ 17 & 0.3525777118 \\ 18 & 0.3525777575 \\ 19 & 0.3525777792 \\ 20 & 0.3525777895 \\ \cdots \\ \infty & 0.3525777990 \end{array} \right)$$

Answer 2

You can work your way out.

You start by writing $g(x, t) $ as a series if powers of $t $ around a point with $x $ as a constant. (The point $t = a $ may be useful)

Now we know

$$\int_a^b g = G(b) - G(a) $$ if $G $ is a primitive of $g $ right?

So just integrate your series for $g$, in order for $t $, term by term to get its primitive. Call that series $S_g$. Because $g $ depended on $x $ and $t $, so does $S_g $ right? Let us keep the order, $S_g(x, t) $. Then compute $S_g(x, f(x)) - S_g(x, a) $. That means we keep $x $ as $x $ and substitute the values of $t$ for the corresponding bounds of the integral. Because the terms will share the exponents you will be able to group them term by term into just one summation and you are done.