Advanced book on Mathematics Olympiad

Real Analysis 537

This we already computed in the previous problem. (“Happiness is longing for repeti-
tion,’’ says M. Kundera.) So the answer to the problem isπ 8 ln 2.
(66th W.L. Putnam Mathematical Competition, 2005, proposed by T. Andreescu)

460.The function lnxis integrable near zero, and the function under the integral sign is
dominated byx−^3 /^2 near infinity; hence the improper integral converges. We first treat
the casea=1. The substitutionx= 1 /tyields

∫∞

0

lnx

x^2 + 1

dx=

∫ 0

∞

ln^1 t

1

t^2 +^1

(

−

1

t^2

)

dt=−

∫∞

0

lnt

t^2 + 1

dt,

which is the same integral but with opposite sign. This shows that fora=1 the integral
is equal to 0. For generalawe compute the integral using the substitutionx=a/tas
follows
∫∞

0

lnx

x^2 +a^2

dx=

∫ 0

∞

lna−lnt

(a

t

) 2

+a^2

·

(

−

a

t^2

)

dt=

1

a

∫∞

0

lna−lnt

1 +t^2

dt

=

lna

a

∫∞

0

dt

t^2 + 1

−

1

a

∫∞

0

lnt

t^2 + 1

dt=

πlna

2 a

.

(P.N. de Souza, J.N. Silva,Berkeley Problems in Mathematics, Springer, 2004)

461.The statement is misleading. There is nothing special about the limits of integration!
Theindefiniteintegral can be computed as follows:

∫

xcosx−sinx

x^2 +sin^2 x

dx=

∫ cosx

x −

sinx

x^2

1 +

(sinx

x

) 2 dx=

∫

1

1 +

(sinx

x

) 2

(

sinx

x

)′

dx

=arctan

(

sinx

x

)

+C.

Therefore,

∫ π 2

0

xcosx−sinx

x^2 +sin^2 x

dx=arctan

2

π

−

π

4

.

(Z. Ahmed)

462.Ifαis a multiple ofπ, thenI(α)= 0. Otherwise, use the substitutionx =
cosα+tsinα. The indefinite integral becomes
∫
sinαdx
1 − 2 xcosα+x^2

=

∫

dt

1 +t^2

=arctant+C.

It follows that the definite integralI(α)has the value