与对数和正弦的确定积分

Steven 06/20/2018. 3 answers, 191 views
integration

通过按部分集成和替换$ x = \ sin t $,我们可以很容易地计算积分$ \ int_ {0} ^ {1} \ ln(x + \ sqrt {1-x ^ 2})dx $等于$ \ sqrt {2} \ ln(\ sqrt {2} +1)-1。$

我试图使用相同的替换$ x = \ sin t $来计算积分$ \ int_ {0} ^ {1} \ frac {\ ln(x + \ sqrt {1-x ^ 2})} {x } dx,$成为

$ \ int_ {0} ^ {\ frac {\ pi} {2}} \ frac {\ ln \ sin(t + \ frac {\ pi} {4})} {\ sin t} dt $

解决特定积分似乎很难。 有帮助吗? 谢谢

3 Answers


ComplexYetTrivial 06/20/2018.

在$ \ frac {1} {\ sqrt {2}} $中拆分积分并在第二部分中使用替换$ x = \ sqrt {1-y ^ 2} $来获得\ begin {align} I&\ equiv \ int \ limits_0 ^ 1 \ frac {\ ln(x + \ sqrt {1-x ^ 2})} {x} \,\ mathrm {d} x = \ int \ limits_0 ^ {\ frac {1} {\ sqrt {2}}} \ frac {\ ln(x + \ sqrt {1-x ^ 2})} {x} \,\ mathrm {d} x + \ int \ limits _ {\ frac {1} {\ sqrt {2 } 1 ^ \ frac {\ ln(x + \ sqrt {1-x ^ 2})} {x} \,\ mathrm {d} x \\&= \ int \ limits_0 ^ {\ frac {1} { \ sqrt {2}}} \ frac {\ ln(x + \ sqrt {1-x ^ 2})} {x} \,\ mathrm {d} x + \ int \ limits_0 ^ {\ frac {1} {\ sqrt {2}}} \ frac {y \ ln(y + \ sqrt {1-y ^ 2})} {1-y ^ 2} \,\ mathrm {d} y = \ int \ limits_0 ^ {\ frac { 1} {\ sqrt {2}}} \ frac {\ ln(x + \ sqrt {1-x ^ 2})} {x(1-x ^ 2)} \,\ mathrm {d} x \ ,. \ end {align}现在让$ x = \ sin(t / 2)$找到\ begin {align} I&= \ frac {1} {2} \ int \ limits_0 ^ {\ frac {\ pi} {2 }} \ frac {\ ln \ left(\ sin \ left(\ frac {t} {2} \ right)+ \ cos \ left(\ frac {t} {2} \ right)\ right)} {\ sin \ left(\ frac {t} {2} \ right)\ cos \ left(\ frac {t} {2} \ right)} \,\ mathrm {d} t = \ frac {1} {2} \ int \ limits_0 ^ {\ frac {\ pi} {2}} \ frac {\ ln \ left [\ left(\ sin \ left(\ frac {t} {2} \ right)+ \ cos \ left(\ frac {吨} {2} \右)\右)^ 2 \右]} {2 \罪\左(\压裂{吨} {2} \右)\ COS \左(\压裂{吨} {2} \右)} \,\ mathrm {d} t \\&= \ frac {1} {2} \ int \ limits_0 ^ {\ frac {\ pi} {2}} \ frac {\ ln \ left(1 + 2 \罪\左(\压裂{吨} {2} \右)\ COS \左(\压裂{吨} {2} \右)\右)} {2 \罪\左(\压裂{吨} {2} \ right)\ cos \ left(\ frac {t} {2} \ right)} \,\ mathrm {d} t = \ frac {1} {2} \ int \ limits_0 ^ {\ frac {\ pi} { 2}} \ frac {\ ln \ left(1+ \ sin(t)\ right)} {\ sin(t)} \,\ mathrm {d} t \\&= \ frac {1} {2} \ int \ limits_0 ^ {\ frac {\ pi} {2}} \ frac {\ ln \ left(1+ \ cos(t)\ right)} {\ cos(t)} \,\ mathrm {d} t \ ,。 \ end {align}定义(来自这个问题的想法)$$ f(a)\ equiv \ frac {1} {2} \ int \ limits_0 ^ {\ frac {\ pi} {2}} \ frac {\ ln \ left(1+ \ cos(a)\ cos(t)\ right)} {\ cos(t)} \,\ mathrm {d} t $$ for $ a \ in [0,\ frac {\ pi} { 2}] $并观察$ f(0)= I $和$ f(\ frac {\ pi} {2})= 0 $。 计算(使用$ \ tan(\ frac {t} {2})= s $)\ begin {align} f'(a)&= - \ frac {\ sin(a)} {2} \ int \ limits_0 ^ {\ frac {\ pi} {2}} \ frac {1} {1+ \ cos(a)\ cos(t)} \,\ mathrm {d} t = - \ sin(a)\ int \ limits_0 ^ 1 \ frac {\ mathrm {d} s} {1+ \ cos(a)+(1- \ cos(a))s ^ 2} \\&= - \ frac {\ sin(a)} {1+ \ cos(a)} \ sqrt {\ frac {1+ \ cos(a)} {1- \ cos(a)}} \ arctan \ left(\ sqrt {\ frac {1- \ cos(a)} { 1+ \ cos(a)}} \ right)\\&= - \ frac {\ sin(a)} {\ sqrt {1- \ cos ^ 2(a)}} \ arctan \ left(\ tan \ left (\ frac {a} {2} \ right)\ right)= - \ frac {a} {2} \ ,. \ end {align}最后,$$ I = f(0)= f \ left(\ frac {\ pi} {2} \ right)+ \ int \ limits _ {\ frac {\ pi} {2}} ^ 0 f'(a)\,\ mathrm {d} a = 0 + \ int \ limits_0 ^ {\ frac {\ pi} {2}} \ frac {a} {2} \,\ mathrm {d} a = \ frac {\ pi ^ 2} {16} \,。$$


Jack D'Aurizio 06/21/2018.

Version 1 。 $$ \ int_ {0} ^ {1} \ frac {\ log(x + \ sqrt {1-x ^ 2})} {x} \,dx = \ int_ {0} ^ {\ pi / 2} \ log (\ sin \ theta + \ cos \ theta)\ cot(\ theta)\,d \ theta \ tag {1} $$通过强制替换$ \ theta \到\ frac {\ pi} {2} - \ theta $平均值相当于$$ \ int_ {0} ^ {\ pi / 2} \ frac {\ log(\ sin \ theta + \ cos \ theta)} {2 \ sin \ theta \ cos \ theta} \ ,d \ theta \ stackrel {\ theta \ mapsto 2 \ arctan u} {=} \ int_ {0} ^ {1} \ frac {\ log(1 + 2t-t ^ 2) - \ log(1 + t ^ 2)} {2t(1-t ^ 4)} \,dt \ tag {2} $$可以通过部分分数分解,通过dilogarithm功能标识 $(3) - (7)$,自$$ \ int \ frac {\ log(1-t)} {t} \,dt = C- \ text {Li} _2(t)。\ tag {3} $$同样适用的是我们避免初始对称化,因为$$ \ int_ {0} ^ {\ pi / 2} \ frac {\ log(\ sin \ theta + \ cos \ theta)} {\ tan \ theta} \,d \ theta = \ int_ {0} ^ {1 } \左[\日志(1 + 2T-吨^ 2) - \日志(1 + T ^ 2)\右] \压裂{1-叔^ 2} {T(1 + T ^ 2)} \,dt的\标签{4} $$


Version 2 。 通过在$(1)$中立即替换$ \ theta = \ arctan u $,原始积分将转换为$$ \ int_ {0} ^ {+ \ infty} \ frac {\ log(1 + u) - \ frac {1} {2} \ log(1 + u ^ 2)} {u(1 + u ^ 2)} \,du $$由Feynman的技巧等于$$ \ int_ {0} ^ {1} \ frac { \ pi + 2a \ log a} {2(1 + a ^ 2)} \,da + \ frac {1} {4} \ int_ {0} ^ {1} \ frac {\ log a} {1-a } \,DA = \压裂{\ PI ^ 2} {8} - \压裂{\ PI ^ 2} {48} - \压裂{\ PI ^ 2} {24} = \颜色{蓝} {\压裂{ \ pi ^ 2} {16}}。\ tag {5} $$ (Poly)对数积分总是一件棘手的事情,事先不知道什么是强制替换或利用某种对称性的最佳时刻。 在这种情况下,通常的切线半角替换只是在简单的解决方案中引入了绕道。


Version 3 。 通过考虑$ \ log \ sin $和$ \ log \ cos $的傅立叶级数,我们有这个,在分布意义上与$ L ^ 2( - \ pi / 2,\ pi / 2)$,$$ \相关cot \ theta = 2 \ sum_ {k \ geq 1} \ sin(2k \ theta)$$ $$ \ log(\ sin \ theta + \ cos \ theta)= - \ frac {\ log 2} {2} - \ sum_ {k \ geq 1} \ frac {\ cos(2k \ theta + k \ pi / 2)} {k} $$因此由Parseval定理$$ \ int_ {0} ^ {\ pi / 2} \ log( \ sin \ theta + \ cos \ theta)\ cot(\ theta)\,d \ theta = \ frac {\ pi} {4} \ sum _ {\ substack {k \ geq 1 \\ k \ text {odd}}} \ frac {( - 1)^ {(k-1)/ 2}} {k} = \ frac {\ pi} {4} \ cdot \ frac {\ pi} {4} = \ color {red} {\ frac {\ pi ^ 2} {16}} \ tag {6} $$ ...哇! 这种方法允许对$ \ int_ {0} ^ {\ pi / 2} \ log(\ sin \ theta + \ cos \ theta)\,\ omega(\ theta)\,d形式的许多积分进行简单而明确的评估。 \ theta $,因此许多积分形式为$ \ int_ {0} ^ {1} \ log(x + \ sqrt {1-x ^ 2})\,w(x)\,dx $。 “向后思考”,原始问题也可能通过计算时刻来解决$ \ int_ {0} ^ {1} x ^ {2m + 1} \ log(x + \ sqrt {1-x ^ 2})\, dx $,然后执行插值/解析延续。


Shashi 06/22/2018.

正如James Arathoon在评论中提到的,通过替换$ t = \ sqrt [] {\ frac {1-x ^ 2} {x ^ 2}} $积分等于:\ begin {align} I:= \ int ^ 1_0 \ frac {\ log(x + \ sqrt [] {1-x ^ 2})} {x} \,dx = \ int ^ \ infty_0 \ frac {t \ log \ left(\ frac {t + 1 } {\ sqrt [] {t ^ 2 + 1}} \ right)} {t ^ 2 + 1} \,dt \ end {align}可以重写一下:\ begin {align} I = \ frac 1 2 \ int ^ \ infty_0 \ frac {t \ log \ left(\ frac {(t + 1)^ 2} {t ^ 2 + 1} \ right)} {t ^ 2 + 1} \,dt = \ frac 1 2 \ int ^ \ infty_0 \ frac {t \ log \ left(1+ \ frac {2t} {t ^ 2 + 1} \ right)} {t ^ 2 + 1} \,dt \ end {align}现在定义以下函数$ F:[0,1] \到\ mathbb R $如下:\ begin {align} F(a):= \ frac 1 2 \ int ^ \ infty_0 \ frac {t \ log \ left( 1+ \ frac {2at} {t ^ 2 + 1} \ right)} {t ^ 2 + 1} \,dt \ end {align}使用Feynman的Trick得到:\ begin {align} F'(a)= \ int ^ \ infty_0 \ frac {t ^ 2} {(t ^ 2 + 1)(t ^ 2 + 2at + 1)} \,dt \ end {align}这个积分不是很难计算,例如一个可以通过部分分数分解或轮廓积分来得到:\ begin {align} F'(a)= \ frac {\ arctan \ left(\ frac {\ sqrt [] {1-a ^ 2}} {a} \钻机 ht)} {2 \ \ sqrt [] {1-a ^ 2}} \ end {align}我们知道:\ begin {align} I = F(1)= \ int ^ 1_0 F'(a)\, da = \ frac {1} {2} \ int ^ 1_0 \ frac {\ arctan \ left(\ frac {\ sqrt [] {1-a ^ 2}} {a} \ right)} {\ sqrt [] { 1-a ^ 2}} \,da \ end {align}这看起来有点可怕,但是在设置$ a = \ cos(x)$后它是非常无辜的,因为那样得到:\ begin {align} I = \ frac {1} {2} \ int ^ 0 _ {\ pi / 2} \ frac {\ arctan \ left(\ tan(x)\ right)} {\ sin(x)}( - \ sin(x) )\,dx = \ frac {1} {2} \ int ^ {\ pi / 2} _0 x \,dx = \ frac {\ pi ^ 2} {16} \ end {align}


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