# Definite integral with Infinite border?

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)?

-
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

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:

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).

-