# Divergent integral error

I'm integrating a product of two functions which is bounded on the interval from -Inf to 0, and diverges from 0 to +Inf. The error that I get says that the integral diverges on {0,+Inf}... How did those bounds get included?

The result[t] function includes a sum of well-defined Hypergeometric functions which are also part of the error message. The approx[t] has some symbolic code in it but it hasn't caused any problems with Integrate in the past.

The integrand looks like this:

The "a" that you see throughout is a symbolic constant. It's got no value associated with it right now.

I see how the integrand should technically diverge when t approaches zero because of the inverse powers, but if we take a look at what the function actually looks like on a plot there doesn't seem to be a problem at t = 0 (that's why I'm hoping the integral will work):

Thanks for any help!

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• Please provide the integrand, so that readers can experiment with it. – bbgodfrey Jul 10 '15 at 19:11
• People here generally like users to post code as Mathematica code instead of images or TeX, so they can copy-paste it. It makes it convenient for them and more likely you will get someone to help you. You may this this meta Q&A helpful – Michael E2 Jul 11 '15 at 4:24

When evaluating an integral, Mathematica applies a series of change-of-variables substitutions in attempt to put this integral in a standard from. Likely ehat has happened here is that Mathematica has performed a substitution like $t\to -t$ which has the effect of changing the bounds of integration. The resulting integral does not converge which almost certainly means that your original integral doe snot converge either. If you have reason to believe that your integral does converge, you could try integrating term by term.
• "integrating term by term" - not always a good idea; consider $$\int_0^\infty \frac{1-\cos x}{x^2} \mathrm dx$$ – J. M.'s technical difficulties Jul 11 '15 at 1:38