# Solving an Improper Integral

66 views (last 30 days)

Show older comments

Hello,

I've tried to solve the following integral as such:

syms x;

f = 1/(sqrt(exp(x)-x))

A = int(f, 1, inf)

and the answer was the function itself:

int(1/(exp(x) - x)^(1/2), x, 1, Inf)

Did I make an error because why is the answer not coming out?

Thank you.

##### 2 Comments

darova
on 8 Apr 2020

Maybe it's too complicated for symbolic calculations. What about numerical solution?

syms x;

f = 1/(sqrt(exp(x)-x))

F = matlabFunction(f);

integral(F,1,inf)

2/sqrt(exp(1))

output

ans =

1.3725

ans =

1.2149

doesn't look like truth

Ameer Hamza
on 8 Apr 2020

### Answers (2)

John D'Errico
on 21 Mar 2021

Think of it like this. Do ALL problems you might pose to a computer have a solution? If so, then I might try

solve('Peace in the middle East')

To no surprise, MATLAB would have a fit if I try it, generating some randomly useless error message.

But many (even most) problems you might pose in mathematics actually have no analytical solution. It is frighteningly easy to pose such a problem. In fact, it looks like you did exactly that.

One solution, and a reason why there are numerical analysis courses taught at many schools, is to learn to use numerical methods to solve intractable problems, where no simple algebraic solution is available. Often numerical analysis is valuable even for problems where a solution exists in theory, yet it is numerically difficult to work with. But this is a probem of the first sort, where no analytical solution apparently exists. At least neither of MATLAB or Wolfram Alpha are able to offer one.

In MATLAB, you can then often tell MATLAB to look for a numerical solution. You can use vpa for that, or you can use vpaintegral.

syms x;

f = 1/(sqrt(exp(x)-x));

int(f,1,inf)

But this will work

vpa(int(f,1,inf))

Or you might do it as

vpaintegral(f,1,inf)

##### 0 Comments

### See Also

### Categories

### Community Treasure Hunt

Find the treasures in MATLAB Central and discover how the community can help you!

Start Hunting!