Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top


I $ = \int_0^{1000} \dfrac{e^{-10x} \sin(x)}{x} dx $

Evaluate I to within $\pm 10^{-5}$

I've broken down the problem to simply evaluating the integral from zero to one (by way of the accuracy required and using a comparison), after that, however, I've no clue how to approach this - I know I could manually take derivatives and somehow discern the taylor series of the integrand, but this is a bad function to derive multiple times. Using a trapezoidal approximation would take way too many sub-intervals - is there another approach to this?

share|cite|improve this question
You want to do it by hand, or program a computer to do it? – icurays1 Apr 4 '13 at 22:04
By hand, but a calculator is permitted. – kvmu Apr 4 '13 at 22:08

First, notice that the expression to be integrated rapidly decreases as x increases. So, you only have to integrate from 0 to some $a$.
You can determine an upper bound for remaining part of the integral by excluding oscillating factor $\sin(x)$ and evaluating $\int_a^{1000}\frac{\exp(-10x)}{x}dx$ (it will require Ei function evaluation - from tables or somehow). Then take good enough $a$ so that this upper bound is less than specified tolerance of $10^{-5}$ and use any usual method such as trapezoid or rectangle approximation to evaluate $\int_0^a\frac{\exp(-10x)\sin(x)}{x}dx$.
From my checks it appeared sufficient to have $a=1$.

share|cite|improve this answer
I've already established that the integral from 0 to 1 is sufficient, however evaluating the integral is the difficulty. Also it needs to be shown that the accuracy is within the desired range. – kvmu Apr 4 '13 at 23:01
@kvmu Well you can just expand $\frac{\sin(x)}{x}$ to series (where series up to $x^4$ term appears sufficient) and integrate symbolically. As for accuracy, I've already provided a means of checking that integral from 1 to 1000 is small enough, you only have to prove that $\text{sinc}(x)$ expansion up to 4th power is enough (e.g. finding $\sup_{x\in [0,1]}{\left(\text{sinc}(x)-\left(1-\frac{x^2}{6}+\frac{x^4}{120}\right)\right)}‌​$ and proving it's smaller than some value. – Ruslan Apr 5 '13 at 7:17
Thanks for the reply! Fortunately, I figured it out before the problem was due - I did something similar to what you describe here: 1) showed that it is sufficient to compute the integral from 0 to 1 2) expanded $\dfrac{sin(x)}{x}$ into a series of partial sums from k = 1 to n + an error term. Since the series is alternating, the |error term| is at least the next term in the series, ie. the n+1th term. Then I multiplied the exponential to the the series and the error term. – kvmu Apr 7 '13 at 3:32
3) Then I did a comparison to show that the |new error term| is at least some value and integrated it. After that, I solved the inequality to show that n = 2 is good enough (for the accuracy required). The integral I ended up solving was something in the form of: $\int_0^1 e^{-10x} -\dfrac{x^2e^{-10x}}{3!}dx$ – kvmu Apr 7 '13 at 3:55

Let $$I(y)=\int_0^y\frac{e^{-10x}\sin x}{x}\,\mathrm dx.$$ Then for $y>0$ we have $$ \left|I(y)-I(\infty)\right|<\frac1y\int_y^\infty e^{-10x}\,\mathrm dx=\frac{e^{-10y}}{10y}.$$ Therefore $|I(1000)-I(1)|\le |I(1000)-I(\infty)|+|I(1)-I(\infty)|\approx 4.54\cdot 10^{-6}$ allows us to merely calculate $I(1)$ if we keep the numerical error below $5\cdot 10^{-6}$.

Note that $0<\frac{\sin x}{x}\le 1$ for $0\le x\le 1$ and $g(x):=\frac{\sin x}{x}$ is strictly decreasing. Therefore, if $0=x_0<x_1<\ldots <x_n=1$, then $$ \sum_{k=1}^n g(x_{k})\int_{x_{k-1}}^{x_k}e^{-10x}\,\mathrm dx<I(1)<\sum_{k=1}^n g(x_{k-1})\int_{x_{k-1}}^{x_k}e^{-10x}\,\mathrm dx,$$ i.e. $$ \sum_{k=1}^n g(x_k)\frac{e^{-10x_{k-1}}-e^{-10x_{k}}}{10}<I(1)<\sum_{k=1}^n g(x_{k-1})\frac{e^{-10x_{k-1}}-e^{-10x_k}}{10}.$$ We need to keep the difference $$ \sum_{k=1}^n (g(x_{k-1})-g(x_k))\frac{e^{-10x_{k-1}}-e^{-10x_{k}}}{10}$$ small, which can be achieved by suitable chice of $x_k$. Note that the sequence can be chosen to grow quite rapidly because of the small eponential factor, except perhaps near $0$. But near $0$, the first factor $g(x_{k-1})-g(x_k)$ is approximately linear in $x_k-x_{k-1}$, for $x_{k-1}=0$ even quadratic.

share|cite|improve this answer
Your approximation is nice, but does not give I to $\pm 10^{-5} $ degree of accuracy (also this needs to be done by hand). For reference: I = 0.0996687 – kvmu Apr 4 '13 at 23:06

I would try to reduce it more, since when $x \gg 0$, we still have $e^{-10x} \ll 1$, and for small $x$, roughly $\sin x \approx x$ so the integrand becomes $e^{-10x}$, which is trivial to integrate.

share|cite|improve this answer
Good call, but I've already checked and it doesn't work. The value of $\int_0^{1} e^{-10x} dx$ is not in the range of the desired accuracy. – kvmu Apr 4 '13 at 23:18

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.