Problems on Improper Integrals

Exercise 1. Decide on the convergence or divergence of

displaymath210

Answer. First notice that the denominator is equal to 0 when x=1. Then the function inside the integral sign is unbounded at x=1. Hence we have two bad points 1 and tex2html_wrap_inline216 . So we must split the integral and write

displaymath218

Let us first take care of the integral

displaymath220

We have

displaymath222

Since (by Bertrand's test) the improper integral

displaymath224

is convergent, then by the comparison test the improper integral

displaymath220

Next we take care of the integral

displaymath228

First notice that when tex2html_wrap_inline230 , then

displaymath232

One may check this by showing that

displaymath234

On the other hand (and this is the crucial step in this exercise) is to find a polynomial approximation of tex2html_wrap_inline236 when tex2html_wrap_inline230 . This will be done via Taylor polynomials. Indeed, we have

displaymath240

when tex2html_wrap_inline230 , which gives

displaymath244

Hence we have

displaymath246

which implies

displaymath248

when tex2html_wrap_inline230 . The p-test implies that the improper integral

displaymath252

is convergent. Therefore the limit test implies that the improper integral

displaymath228

is convergent. Putting the two integrals together, we conclude that the improper integral

displaymath210

is convergent.

Exercise 2. Decide on the convergence or divergence of

displaymath258

Answer. The term tex2html_wrap_inline260 is never equal for tex2html_wrap_inline262 . So let us focus on the term tex2html_wrap_inline264 . According to the domain of the tangent function, the only bad points we have to worry about is tex2html_wrap_inline266 . Clearly we have

displaymath268

Hence we have an improper integral or Type I at the bad point tex2html_wrap_inline266 . Clearly we have

displaymath272

when tex2html_wrap_inline274 . So let us approximate tex2html_wrap_inline276 when tex2html_wrap_inline274 . Again we will use Taylor polynomials, we have

displaymath280

when tex2html_wrap_inline274 . This gives

displaymath284

when tex2html_wrap_inline274 . Putting the stuff together we get

displaymath288

The p-test implies that the improper integral

displaymath290

is divergent. Therefore the improper integral

displaymath258

is divergent.

Exercise 3. Decide on the convergence or divergence of

displaymath294

Answer. The only improper behavior is around tex2html_wrap_inline216 . Hence this integral is of Type II not of Type I. Therefore no need for splitting it. Note that when tex2html_wrap_inline298 , then tex2html_wrap_inline300 . Hence

displaymath302

when tex2html_wrap_inline298 . The p-test implies that the improper integral

displaymath306

is convergent. Therefore the limit test implies that the improper integral

displaymath294

is convergent.

Problems

Next you will find some not so easy problems on improper integrals. We invite you to solve them and submit the answer to SOS MATHematics. We will publish your answer with your name. Good luck.

Problem 1. First decide on the convergence and divergence of

displaymath310

Then evaluate f(x).

Problem 2. Assume that tex2html_wrap_inline314 is continuous. Find tex2html_wrap_inline316 , where

displaymath318

Problem 3. Consider the function

displaymath320

Find f(x).

Problem 4. Evaluate

displaymath324

Problem 5. In this problem, we will evaluate

displaymath326

1
Evaluate

displaymath328

2
Let tex2html_wrap_inline330 which tex2html_wrap_inline332 function (that is f is differentiable and f' is continuous). Show that

displaymath338

3
Evaluate

displaymath326

Problem 6. Decide on convergence or divergence of

displaymath342

where x > 0 and tex2html_wrap_inline346 is any real number.

Problem 7. Decide on convergence or divergence of

displaymath348

Problem 8. Decide on convergence or divergence of

displaymath350

If I is convergent, evaluate it.

Problem 9. Find

displaymath354

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Mohamed A. Khamsi
Tue Dec 3 17:39:00 MST 1996

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