Here are some challenging integrals to evaluate.

Evaluate . (MIT Integration Bee)

Evaluate .

For , prove that .

If is a bounded non-negative function, then show that .

Evaluate .

Evaluate.

The problems above aren’t necessarily in increasing order of difficulty; however, the last one can be almost impossible to evaluate if one doesn’t know the right “trick”, which will be the subject of my third identity in my series of posts titled, *A few useful identities related to definite integrals*, which you can find in the Problem Corner.

*Note*: Most of the problems (posted above) have been borrowed from people who posted the same in AoPS.

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January 31, 2008 at 6:30 am

A few useful identities related to definite integrals (part 3) « Vishal Lama’s blog[…] apply the above identity to the “difficult” integral in problem from the Integration Bee, Challenging Integrals post to evaluate the integral. The solution turns out to be an easy one. The answer is , just in […]

April 10, 2008 at 1:25 pm

Todd TrimbleSince Vishal has been doing such a good job responding to the various problems and exercises in my recent posts, I thought it’d be only fair if I responded to some of his. 🙂

I can’t claim to have great insight into these problems. But here are a few light hints for some of them.

(1) Chip away at the exponent on the sine by repeatedly using the identities

.

This leads to a nice recurrence from which an answer can be extracted.

(3) Consider substitution of hyperbolic functions (, etc.). For some reason hyperbolic functions don’t seem to be much emphasized in the calculus courses here in the US; here the reflex reaction to this problem would be to substitute a trigonometric function (), but solving it with hyperbolic functions is much cleaner. Most students seem to engage hyperbolic functions seriously only when they take a course in relativity theory.

(4) This is the problem I found easiest. Consider what happens when you substitute . Although I’m a little uneasy about the formulation; if for example f is the constant function 1, then convergence of the improper integral would seem to be an issue. At the risk of giving away the solution, it could be reformulated by taking the bounds of integration to be 1/t and t, and letting t approach infinity. Or, just suppose that f(x) tends to 0 moderately quickly as x approaches infinity — that would work too.

April 11, 2008 at 9:43 pm

Todd TrimbleHmm… having some trouble with number (2).

[Number (5) doesn’t seem so bad: just an integration by parts plus a little rationalizing of denominators.]

I thought at first that (2) would succumb to a “differentiation under the integral sign” trick. Feynman said he loved that trick and used it over and over again (to compute integrals where other methods like contour integration failed), but if that applies here, then I’m not seeing it. Actually, I’ve tried a number of things, to no avail.

Grrr…

August 23, 2012 at 2:05 pm

AryanFor (2), wouldn’t a simple power rule method work? u=arctan x, du=dx/(x+1) [This applies the inverse trigonometric function of int du/(a^2+u^2)=(1/a)arctan(u/a)]

April 11, 2008 at 10:00 pm

Vishal LamaAh! The padawan has finally succeeded in stumping the Jedi Master! 😀 … Just kidding!

On (5), I think it would be a lot easier if we used the substitution .

And now, (2) is special! 🙂 What if we write as and then use integration by parts. This will yield the integral , which has been evaluated here!

April 11, 2008 at 10:26 pm

Todd TrimbleThanks! I did consider, for (2), the integration by parts followed by the substitution , but I wasn’t inspired enough to try those other machinations. Now that I see them, I don’t feel too bad for missing them. 🙂

Now, < wringing hands> I must plot my revenge… bwah ha ha ha!

March 11, 2009 at 11:24 pm

ShanazDid anyone get #6?

March 12, 2009 at 12:52 am

Vishal LamaShanaz,

For #6, you will need to use identity (1) that I mentioned here. If you still can’t figure out the answer, let me know and I would be glad to help you.

March 12, 2009 at 4:10 am

ShanazThanks so much Vishal

March 12, 2009 at 5:44 pm

ShanazVishal; Can you do me the biggest favour and help me for #6 and can you do it step by step

? I find it extreamly hard..

Thanks

March 12, 2009 at 6:01 pm

Vishal LamaShanaz,

I must first apologize to you for providing the wrong hint for #6. Here is the correct hint that I should have given you before. That should help you compute the definite integral. Trust me, it is not hard at all! 🙂

If you still feel that I should post the solution, please don’t hesitate to let me know.

[

Update. Well, after you use the hint that I provided just above, you also do need to use the hint I provided in my last comment to compute the integral.]March 12, 2009 at 10:25 pm

ShanazOk it’s stilll hard. Can you help me pleasee I give uppp…:(

March 12, 2009 at 11:02 pm

Vishal LamaShanaz,

Here is the solution.

(#6)Let us denote the given definite integral by . Then applying identity (3) given here to , we get, which after a little manipulation yields

, which yields

.

Further, applying identity (1) given here to the above equation, we obtain

.

Now adding the above two equations, we get

.

Hence, , which is our answer.

Dang, that was long! You owe me big time for the effort I put into writing the solution. Just kidding! 🙂

[

UpdateOne should note that the integer in the above integral can be replaced by any positive even integer and that can be replaced by any real. The final answer won’t change!]March 12, 2009 at 11:19 pm

Shanazawww thank youuuu so much times 100…lol

April 16, 2009 at 6:51 am

mzargarSolution for 1: Clearly, . Let . Factoring out the denominator, and gives . Substituting gives . Factoring out and dividin and multiplying by , .

April 16, 2009 at 6:30 pm

mzargarComment for 2: Instead of Integration by parts, you can also do the following:

August 31, 2009 at 12:14 pm

Ragib ZamanThe easiest way I can see for (4) is u= ln(x). Comes out very quickly.

January 2, 2011 at 6:28 am

2010 in review « Todd and Vishal’s blog[…] Integration Bee, Challenging Integrals January 2008 16 comments […]

February 15, 2011 at 2:56 am

ChandrasekharI used to love doing integration problems when i was t high school. Here is challenging one: (

Slightly edited.)October 24, 2013 at 6:50 am

Satyajit Nandyplz leave the solution… my result is different….

October 24, 2011 at 6:32 pm

kaggwa jamesits ma first time using dis website but its really cool

January 28, 2012 at 9:32 am

definiteintegralsNice Questions.

You can also solve,

June 23, 2012 at 11:39 pm

Torh Hinnehcan somebody please evaluate this integral for me..

[

Edited latex code -ed.]June 23, 2012 at 11:44 pm

Torh Hinnehwhen i use integration by parts,I obtain the same question as an answer.

[Edited latex code – ed.]

June 23, 2012 at 11:52 pm

Vishal LamaI don’t think the integral has an “elementary” solution.

June 27, 2012 at 9:48 pm

Torh Hinnehcan you please help me with the “nonelementary” solution if any.

June 27, 2012 at 10:44 pm

Torh K. Hinnehor maybe you can just give me a hint to tackle it.

September 15, 2012 at 9:06 am

MehboobHi Torh K. Hinneh,

I think after first integration by parts (using ln(x) as the first function), you can rewrite the exponential in the second term as an infinite sum series. Then the integral can be evaluated easily.

October 1, 2013 at 1:00 pm

dipanjan chatterjeethe answere of problem no. 20 is 1/a 3.b 3

August 18, 2016 at 7:29 pm

Krunal kapadiaCan you answer me above questions with method

December 8, 2016 at 11:16 am

DevSecond question is pretty easy,

Put logx=t

1/x.dx =dt

Limit changes to -infinity to +infinity