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This integral was introduced to me by one of my students. It is from some hard maths contest – I could guess from William Lowell Putnam Mathematical Competition. Anyhow, it took me about one and a half hours to figure this one out, and then some hours to redo it and verify all steps and so on – about four hours all in all.

Ok, so the integral is

It seems like no standard methods would work. One way is to use the trick introduced in the first integral in this section, and that is to insert another variable. This enables us to get rid of the logarithm. Let

The reason for putting the variable there is that it will enable us to get rid of the logarithm, as we will see.

**Differentiate with respect to p**

Then we differentiate with respect to *p* to get

I.e.

Here we have to remember that *p* is our variable and that we need to use the chain rule.** **

**Partial fractions**

To solve the integral we need to rewrite our integrand using partial fractions

We should find values for *A*, *B* and *C* such that

The reason for the numerator *Bx*+C is that, when we multiply away the denominators, the two terms in the right-hand side (RHS) should be of the same degree – as we will see is necessary.

Next, we multiply both sides with the denominator of the left-hand side (LHS) to get

Now comes a really neat trick – so if you have not worked with partial fractions then listen carefully (as I would have said if this would have been a lesson).

When *x* varies the constant part of the LHS and the RHS must be the same. The same with the part that varies with *x* and the part that varies with *x*^{2}. Each of these parts must vary in unison with each other. We thus get

Here we just write the coefficient, since we don´t care about any particular value of *x*. From the first equation, we get that *A*= –*C*. This gives us

We may now multiply the first equation of this with *p* to get

Next, we subtract the second equation from the first to get

or

From this, and that *A*= –*C* we get

Finally, from

We get

Back to our rational function we thus get

One can check that this is correct by adding the fractions (find a common denominator) and then by expanding and simplification.** **

**Back to the integral**

We now have

** **

**Solving the three integrals**

For the first integral, we get

The second one is

To solve this we can make the substitutions

to get

The third one gives us

Combining this we get

**Undoing the Differentiation with respect to p**

Ok, to get back to *I*(*p*) we need to take the anti-derivative of *I’*(*p*).

The second term is virtually the same as the third one above, so we get

(We wait with the constant of integration.)

The third term is like the second one above, and we have

The third integral may be the most difficult to figure out, but if we remember that

then we can rewrite the integral as

**Putting it all together**

This gives us, after adding the constant of integration we get from the indefinite integrals

Solving this for a value of p that we can solve it for, namely *p*=0 we get

or

So *C*=0.

Now we can let *p*=1 to get back to our original integral

or

It is rather fascinating to see that the two last terms turned out to be the same. That gives us

We have thus basically been able to turn the integral to itself – but happily enough with a minus sign in front of it. We thus have

or

This finally gives us

Phew. That was a tough one.

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