Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

I am instructed to use definite integral to calculate things such as $\int_{a=0}^{b=\infty}x^{7}$ but $\Delta x=\frac{a-b}{n}$ diverges. PatrickJMT shows here to use the border-difference divided by n for $\Delta x$, what should I use now? Can I get the integral somehow to definite form for example by substitution or some trick or perhaps the goal of this question is to challenge people (not everything definite but indefinite)?

share|improve this question
2  
Integrals such as $\int_a^\infty f(x)\ dx$ are called improper integrals. The meaning is that this is the limit (if it exists) as $b \to \infty$ of the definite integral $\int_a^b f(x)\ dx$. In this case $\int_0^b x^7\ dx = b^8/8$, and the limit of that as $b \to \infty$ is $+\infty$. We would therefore say that this improper integral diverges. –  Robert Israel Feb 5 '12 at 8:27
add comment

1 Answer 1

up vote 1 down vote accepted

As mentioned in the comments, this is what is called an improper integral. You do not want to fiddle around with $\Delta x$'s or set up Riemann sums here. Instead, consider the integral as an area: the definite integral of the nonnegative function $f$ $$ \int_1^\infty f(x)\, dx $$ is the area under the graph of $f$ over the interval $[1,\infty)$-the area of the blue and red regions below:

enter image description here

However, this area may be infinite, the area of the red piece may be infinite in particular. On the other hand, the $f$ pictured above is asymptotic to the $x$-axis, so perhaps the area of the red piece becomes small as we take $b$ large.

Formally what we do is calculate $$ \int_1^b f(x)\, dx $$ first. This gives the area of the blue piece.

Then we examine what happens as $b$ becomes large. We calculate $$ \lim_{b\rightarrow\infty} \int_1^b f(x)\, dx $$

If this limit exists and is equal to $L$, we then write $$ \int_1^\infty f(x)\, dx=L $$ and say the improper integral converges to $L$.

If the limit does not exist, we say $ \int_1^\infty f(x)\, dx $ diverges.


For example, consider $$ \int_1^\infty {1\over x^2}\,dx. $$

Calculate:

$\ \ \ \int_1^b {1\over x^2}\,dx = (-x^{-1})|_1^b= 1-{1\over b}$.

then calculate

$\ \ \ \lim\limits_{b\rightarrow\infty} \bigl[ 1-{1\over b}\bigr]=1$.

So $$ \int_1^\infty{1\over x^2}\,dx= \lim\limits_{b\rightarrow\infty}\int_1^b {1\over x^2}\,dx =1. $$ The improper integral $\int_1^\infty {1\over x^2}\,dx $ converges to 1.


Integrals of this type do not always converge. The "area" shown above may be infinite. You can verify this by computing $\int_1^\infty{1\over x}\,dx$.

The example in your question, $\int_1^\infty x^7 \,dx$ , gives a divergent integral also. (Draw the graph of $y=x^7$ and this should be obvious here).

share|improve this answer
add comment

Your Answer

 
discard

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.